UNIVERSITY  OF  CALIFORNIA 
AT  LOS  ANGELES 


GIFT  OF 

Dr.   ERNEST   C.   MOORE 


TEACHING  ARITHMETIC 

A  HANDBOOK   FOR   TEACHERS 

AND 

A  TEXTBOOK  FOR  NORMAL  AND 
TRAINING  SCHOOLS 


BY 


MIDDLESEX   A.   BAILEY,   A.M. 

HEAD  OP  THE    DEPARTMENT   OP   MATHEMATICS   OF   THE    NEW  YORK 
TRAINING    SCHOOL   FOR    TEACHERS,    NEW  YORK    CITY 


PUBLISHED   BY 

MIDDLESEX    A.    BAILEY 

YONKERS,  N.  Y. 


OOPYEIGHT,    1913,   BY 

MIDDLESEX  A.  BAILEY. 


COPYRIGHT,  1913,  IN  GREAT  BRITAIN. 


J.  8.  Gushing  Co.  —  Berwick  &  Smith  Co. 
Norwood,  Mass.,  U.S.A. 


"55 


PREFACE 

THERE  is  a  science  of  teaching  arithmetic.  The  feeling 
of  needs  and  the  finding  of  means  for  their  satisfaction  is 
the  law  of  advancement  ;  logical  division  is  the  law  of 
classifications  and  definitions  ;  induction,  deduction,  and 
the  complete  method  are  the  laws  for  the  discovery  of 
principles;  experimentation  and  reasoning  are  the  laws 
for  the  solution  of  simple  problems  ;  and  the  analysis  of 
complex  problems  into  simple  problems  is  the  law  for  the 
solution  of  problems  in  general. 

In  Part  I  the  author  has  endeavored  to  develop  and 
state  these  laws.  In  Part  II  he  has  endeavored  to  illus- 
trate their  application  ;  much  has  been  omitted  but 
enough  has  been  given  to  suggest  the  rest.  In  Part  III 
he  has  endeavored  to  give  an  idea  of  the  advance  in 
complexity  from  grade  to  grade  in  elementary  schools  and 
of  what  is  required  to  secure  primary  and  higher  licenses 
for  teaching  arithmetic. 

As  a  handbook,  this  work  should  help  the  superintend- 
ent to  secure  uniformity  in  his  schools  ;  the  principal  to 
harmonize  the  work  of  the  grades  ;  and  the  teacher  to  get 
a  clear  view  of  fundamental  laws. 

As  a  textbook,  this  work  should  help  the  teacher  in 
normal  and  training  schools  to  supplement  his  lectures  ; 
and  prospective  teachers  in  school  and  out  of  school  to 
prepare  for  their  chosen  profession. 

MIDDLESEX  A.  BAILEY. 

NEW  YORK  TRAINING  SCHOOL  FOR  TEACHERS, 
NEW  YORK  CITY. 

iii 

219364 


CONTENTS 


PAET  I.     INTRODUCTION 


PAGE 

LESSON    1. 

NEEDS  —  SPECIES  —  LOGICAL  DIVISION     r  . 

2 

LESSON    2. 

CASES  —  LOGICAL  DIVISION  —  MEASUREMENTS    . 

6 

LESSOX    3. 

ONE-LINE  DIAGRAMS.  —  MECHANICAL  AIDS 

10 

LESSON    4. 

PRINCIPLES  —  INDUCTION  —  DEDUCTION 

14 

LESSON    5. 

PRINCIPLES  —  COMPLETE  METHOD 

18 

LESSON     6. 

SIMPLE  PROBLEMS  —  BY  EXPERIMENT 

22 

LESSON    7. 

SIMPLE  PROBLEMS  —  BY  REASONS 

26 

LESSON    8. 

COMPLEX  PROBLEMS  —  BY  ARITHMETIC 

30 

LESSON    9. 

WRITTEN  PROBLEMS  —  ARRANGEMENT 

34 

LESSON  10. 

PROBLEMS  —  BY  ALGEBRA  —  BY  FORMULA  —  BY 

RULE  —  BY  PROPORTION          .... 

38 

LESSON  11. 

PROBLEMS  —  BY  PROPORTION  —  BY  VARIATION  . 

42 

LESSON  12. 

LESSON  PLANS        

46 

LESSON  13. 

IN  GENERAL    

50 

PART   II.     SUBJECT   MATTER 

LESSON  14. 

NOTATION  AND  NUMERATION        .... 

55 

LESSON  15. 

NOTATION  AND  NUMERATION        .... 

60 

LESSON  16. 

ADDITION         

64 

LESSON  17. 

SUBTRACTION  

71 

LESSON  18. 

MULTIPLICATION     

76 

LESSON  19. 

DIVISION          

82 

LESSON  20. 

FACTORING      

90 

LESSON  21. 

96 

LESSON  22. 

DECIMALS        

103 

iv 

CONTENTS 


LESSON  23. 

DENOMINATE  NUMBERS          .... 

PAGE 

.     110 

LESSON  24. 

MENSURATION         

.    117 

LESSON  25. 

INVOLUTION,  EVOLUTION,  LOGARITION 

.     125 

LESSON  26. 

ALGEBRA  IN  ARITHMETIC     .... 

.     131 

LESSON  27. 
LESSON  28. 

PERCENTAGE   .        .        ... 

.     137 
144 

LESSON  29. 

INTEREST          

.     152 

LESSON  30. 

INTEREST         

.     160 

PAKT   III.     EXERCISES 


SECTION  1.  ELEMENTARY  SCHOOLS  . 

SECTION  2.  PRIMARY  LICENSE  —  CITY 

SECTION  3.  PRIMARY  LICENSE  —  STATE 

SECTION  4.  HIGHER  LICENSES 


169 
181 
187 
190 


TEACHING  ARITHMETIC 

PART   I.     INTRODUCTION 

1.  Object  and  Scope.     The  office  of  this  book  is  to  dis- 
cuss means  of  assisting  pupils  six  years  of  age  to  pass  in 
eight  years  from  complete  ignorance  of  mathematics  to 
the  knowledge  and  efficiency  outlined  in  the  .course  of 
study.     The  keynote  is,  Every  step  in  response  to  a  need 
and  as  a  means  to  an  end. 

The  race  has  advanced  to  its  present  stage  and  will  ad- 
vance to  other  stages  in  response  to  needs.  The  satisfac- 
tion of  one  need  gives  rise  to  others,  their  satisfaction  to 
still  others,  and  so  on.  Pupils  must  follow  the  same  path 
but  must  accomplish  in  a  few  years  what  the  race  has 
accomplished  in  many  centuries. 

2.  The  Plan.     The  subject  matter   of  mathematics  is 
mental  products  —  number  and  the  operations  upon  num- 
ber do  not  exist  ready  formed  in  nature.     Assuming  that 
the  pupil,  like  the  race,  should  be  an  inventor,  we  shall 
discuss  the  principal  means  of  helping  him  to  create  sub- 
ject matter  for  himself,  and  shall  apply  these  means  to  the 
divisions  of  arithmetic. 

l 


LESSON  1.     NEEDS 

3.  Discovering  Needs.     Pupils  are  to  take  every  step 
in  response  to  a  need. 

Place  them  in  the  actual  situations  that  give  rise  to  a 
need  or  cause  them  to  image  these  situations,  and  suggest 
the  need  by  a  question.  No  labored  effort  should  be 
made  ;  the  suggestion  should  be  natural  and  simple. 

ILL.  Number.  T.  "  On  your  desk  there  is  a  handful  of  sticks. 
How  many  are  there  ?  " 

ILL.  Interest.  T.  "  A  man  borrowed  $  1000.  With  this  money 
he  bought  goods  which  he  sold  for  $  1200.  What  should  he  be  will- 
ing to  do  for  the  money  borrowed  ?  " 

4.  Discovering   Species.     Most    terms    have    varieties. 
Thus,  numbers  are  abstract  or  concrete,  concrete  numbers 
are  denominate  or  not  denominate,  denominate  numbers 
are  simple  or  compound.      We  do  not   wish   pupils   to 
study  these  varieties  ready  formed  and  to  memorize  set 
definitions,  but  we  wish   them  to  find  the  varieties  for 
themselves  and  to  make  definitions  in  terms  of  the  devel- 
opment.     The   process  of   discovering   the  species  of  a 
genus  is  logical  division. 

Fix  upon  a  basis  of  classification,  find  the  differences 
of  this  basis,  unite  the  genus  and  the  differences,  name 
the  resulting  species,  define  the  species. 

Basis.  The  basis  of  classification  is  determined  by 
some  need.  Thus,  we  may  wish  to  find  what  kinds  of 
triangles  there  may  be  with  reference  to  the  relative 
lengths  of  the  sides. 

2 


§4  LESSON   1.    SPECIES  — LOGICAL  DIVISION  3 

Differences.  The  differences  of  a  basis  of  classification 
may  be  found  by  experiment  or  by  the  law,  a  thing  must 
be  or  must  not  be.  Thus,  to  find  the  differences  in  rela- 
tive lengths  of  three  lines  we  may  draw  several  sets  of 
three  lines  and  examine  all  possible  cases.  Or  we  may 
reason,  Three  are  equal  or  not  three  equal ;  of  three  not 
equal,  two  are  equal  or  not  two  equal.  This  gives  rise  to 
three  equal,  two  equal,  or  no  two  equal. 

Names.  Names  are  common  words  selected  for  their 
appropriateness,  or  unusual  words  from  foreign  languages. 
The  pupil  is  helped  to  fix  a  name  in  mind  by  an  explana- 
tion of  its  origin  provided  the  explanation  is  within  his 
comprehension.  Thus,  when  no  two  sides  of  a  triangle 
are  equal,  the  sides  if  placed  parallel  are  like  the  rungs  of 
a  ladder.  The  triangle  is  called  scalene  from  the  Latin 
for  ladder. 

Definitions.  To  define  a  term  logically  is  to  state  its 
genus  and  differences.  There  are  as  many  differences  as 
there  have  been  bases  of  classification.  The  proximate 
genus  should  be  used  when  its  meaning  is  understood 
because  there  is  then  only  one  difference.  Thus,  from 
the  proximate  genus,  A  scalene  triangle  is  a  triangle  hav- 
ing no  two  of  its  sides  equal;  from  a  remote  genus,  A 
scalene  triangle  is  a  polygon  having  three  sides  and  having 
no  two  of  its  sides  equal. 

Teaching.  It  would  never  do  to  speak  to  children 
about  bases  of  classification,  differences  of  the  basis,  genus, 
and  species.  The  formal  steps  are  to  be  in  the  mind  of 
the  teacher  as  a  guide,  the  thought  being  that  the  teacher 
with  such  a  guide  will  do  a  better  piece  of  work  than 
without  it. 


4  LESSON   1.    SPECIES  — LOGICAL  DIVISION  §4 

ILL.  Triangles.  T.  "  Let  us  find  the  different  kinds  of  triangles 
with  reference  to  the  relative  lengths  of  the  sides. 

"  Draw  a  triangle  with  all  its  sides  equal.  Draw  some  other  kind. 
John  has  a  triangle  with  two  sides  equal,  and  James  has  a  triangle 
with  no  two  sides  equal.  Is  any  other  kind  possible  ?  Draw  a  tri- 
angle with  three  sides  equal  and  name  it  equilateral  triangle,  draw  a 
triangle  with  two  sides  equal  and  name  it  isosceles  triangle,  draw  a 
triangle  with  no  two  sides  equal  and  name  it  scalene  triangle.  Define 
each  kind." 

A  /  Equilateral 

Aj^..A  Triangle  (-  -  Isosceles 

.  \  Scalene 

ILL.  Fractions.  T.  "  Let  us  find  the  different  kinds  of  fractions 
with  reference  to  the  relative  value  of  numerator  and  denominator. 

il  Write  a  fraction.  Mary,  how  does  the  numerator  of  your  fraction 
compare  with  its  denominator?  The  numerator  is  less  than  the 
denominator.  This  is  true  of  every  fraction  that  has  been  written. 
Write  some  other  kind.  John  has  a  fraction  whose  numerator  is 
equal  to  its  denominator,  and  Henry  has  a  fraction  whose  numerator 
is  greater  than  its  denominator.  See  if  you  can  find  some  other 
kind.  No  one  succeeds.  We  will  call  such  fractions  as  f  proper. 
The  name  is  appropriate  because  we  can  separate  a  unit  into  3  equal 
parts  and  take  2  of  them.  Define  a  proper  fraction.  We  will  call 
such  fractions  as  f  improper.  After  we  have  separated  a  unit  into  3 
equal  parts  we  cannot  take  4  of  them.  Hence,  f  is  not  properly  a 
fraction.  We  will  call  such  fractions  as  f  improper.  After  we  have 
separated  a  unit  into  3  equal  parts  we  can  take  the  3  parts,  but '  frac- 
tion '  means  a  part  and  '  f '  is  a  whole.  Hence  |  is  appropriately  called 
an  improper  fraction.  Define  an  improper  fraction.  An  improper 
fraction  is  a  fraction  whose  numerator  is  equal  to  or  greater  than  its 
denominator." 

/  Nu.  less  than  den.  /  Proper 

Fraction  (-  -  Nu.  equal  to  den.  Fraction  {-  -  Improper 

\  Nu.  greater  than  den.  N  Improper 


§5  LESSON   1.    SPECIES  — LOGICAL  DIVISION  5 

5.  Diagrams.     At  the  end  of  a  classification,  a  diagram 
should  be  made  to  relate  the  genus  and  the  species.     The 
diagram  may  emphasize  the  differences,  the  finished  prod- 
ucts, or  the  names.     Following  are  diagrams  illustrating 
the  classification  of  quadrilaterals  : 

/Adjacent  sides  equal 
,  All  angles  right  <    Adj   gides  not        ^ 

x  Both  pair  sides  ||       /  ... 

/  \  /Adjacent  sides  equal 

/  NNot  all  right        \Adj.  sides  not  equal 

Quad.  <• One  pair  sides  || 

Neither  pair  sides  || 

/Square 

/Rectangle  <    -T, 

_      ..  ,  /  X  Oblong 

Parallelogram  ^ 

\TM,      u  -j  /Rhombus 
/  Rhomboid  <    ,.T 

,    /  \No  name 

Quad .  c Trapezoid 

Trapezium 

6.  Exercises.     1.   Cause    pupils    to    feel    the    need   of    addition. 
2.   Cause  pupils  to  feel  the  need  of  bills.     3.   Help  pupils  to  classify 
integers   with  reference  to  divisibility  by   2.      4.    Help  pupils  to 
classify  number  with  reference  to  the  expression  of  the  unit.    6.   Ex- 
plain the  appropriateness  of  equilateral  as  the  name  of  a  triangle 
whose  sides  are  all  equal.     6.   Criticise  this  definition :    An  equilat- 
eral triangle  is  a  three-sided  triangle   having  its  sides  all  equal. 

7.  Criticise  this  definition  :    A  triangle  is  one  which  has  three  sides. 

8.  Teach  the  classification  of  quadrilaterals.     9.   Define  square,  using 
as  genus :  (a)  rectangle ;  (6)  quadrilateral. 


LESSON  2.    CASES  —  LOGICAL  DIVISION 

7.  Discovering  Cases.  Different  types  of  problems  may 
involve  the  same  terms.  Thus  in  interest,  the  prin- 
cipal, time,  and  rate  may  be  given  to  find  the  interest ; 
the  principal,  time,  and  interest  may  be  given  to  find  the 
rate  ;  and  so  on.  Pupils  should  discover  the  different 
cases  for  themselves.  The  cases  are  not  usually  named. 

ILL.  Cost  of  1,  En.  Cost,  Number.  T.  "At  3  ^  each  the  cost  of 
5  apples  is  15^.  Let  us  find  the  different  problems  that  can  be 
formed  by  the  omission  of  each  term  in  succession. 

"  State  the  problem  which  arises  from  the  omission  of  15  j*.  At  3  ^ 
each  what  is  the  cost  of  5  apples  V  State  the  problem  which  arises 
from  the  omission  of  3  ?.  If  the  cost  of  5  apples  is  15^,  what  is  the 
cost  of  1  apple?  State  the  problem  which  arises  from  the  omission 
of  5  apples.  At  3  p  each  how  many  apples  can  be  bought  for  15  ^  ?  " 

ILL.  Whole,  Fraction  of  Whole,  Part.  T.  "  Let  us  find  the  differ- 
ent problems  which  arise  from  f  of  20  =  8  by  the  omission  of  each 
term  in  succession. 

"  State  the  problem  which  results  from  the  omission  of  8.  What  is 
f  of  20?  State  the  problem  which  results  from  the  omission  of  20. 
8  is  f  of  what  number  ?  State  the  problem  which  results  from  the 
omission  of  §.  8  is  what  part  of  20?  State  these  problems  in  gen- 
eral form." 

1.  To  find  a  fractional  part  of  a  number. 

2.  To  find  a  number  when  a  fractional  part  is  given. 

3.  To  find  what  fractional  part  one  number  is  of  another. 

The  teacher  should  carry  such  developments  farther 
than  he  proposes  to  carry  them  with  pupils,  for  the  sake 
of  widening  his  own  horizon,  for  he  should  know  more 
than  he  attempts  to  teach. 

6 


§7  LESSON   2.    CASES  — LOGICAL  DIVISION  7 

ILL.  Cases  in  Percentage.  Let  us  find  all  the  different  cases 
in  percentage  from  the  formulae,  P  =  B  x  R,  A  =  B  +  P,  and  D  = 
B-P. 

There  are  three  equations  with  five  quantities.  To  solve  these 
equations  two  of  the  quantities  must  be  known,  because  it  is  impossi- 
ble to  solve  three  equations  with  more  than  three  unknown  quanti- 
ties. Hence,  there  will  be  three  cases  in  percentage  for  every  two 
known  terms. 

The  combinations  of  twos  in  the  terms  A,  B,  D,  P,  R  are :  AB, 
AD,AP,AR;  BD,  BP,  BR ;  DP,  DR;  PR.  That  is,  there  are  10 
times  3,  or  30,  cases  in  percentage.  They  are : 

Given  To  Find                          Given  To  Find 

A  and  B;  D,  P,  R  fiandP;  A,  D,  R 

A  and  D ;  B,  P,  R  B  and  R ;  A,  D,  P 

A  and  P ;  B,  D,  R  D  and  P ;  A,  B,  R 

A  and  R ;  B,  D,  P  D  and  R ;  A,  B,  P 

SandD;  A,  P,  R  P  and  R;  A,  B,  D 

The  teacher  should  form  a  concrete  problem  for  each 
case  grouped  around  some  industry  or  activity. 

ILL.     Sheep.    Let  us  take  the  activity  of  buying  and  selling  sheep. 

A  represents  the  number  of  sheep  after  a  purchase  or  53 ;  B,  the 
original  number  or  50;  Z>,  the  number  after  a  sale  or  47;  P,  the 
number  bought  or  sold  or  3 ;  R,  the  per  cent  of  the  original  number 
or  6%. 

Case  1.  Given  A  and  B  to  find  D.  A  man  had  50  sheep ;  after 
purchasing  a  certain  number  he  had  53.  If  he  had  sold  as  many  as 
he  bought,  how  many  would  he  have  had  left? 

Case  6.  Given  A  and  D  to  find  R.  After  purchasing  a  number 
of  sheep  a  man  had  53  ;  if  he  had  sold  as  many  as  he  purchased,  he 
would  have  had  47  left.  What  per  cent  of  the  original  number  did 
he  purchase  ? 

Case  10.  Given  *4  and  R  to  find  B.  After  purchasing  a  number 
of  sheep  a  man  had  53,  which  was  6  %  more  than  the  original  number. 
What  was  the  original  number? 


8  LESSON  2.    MEASUREMENTS  §8 

8.  Progressive   Difficulty.     The   teacher   must   classify 
the  examples  under  each  subject  in  the  order  of  their 
difficulty   that   the   pupils   may   advance  in   the  line  of 
least  resistance.      He  may  need  several  bases  of  classi- 
fication. 

ILL.  Subtraction  of  Mixed  Numbers.  The  difficulty  depends  upon 
the  sameness  of  the  denominators,  and  upon  the  relative  value  of 
the  fractions  in  the  minuend  and  subtrahend.  Classifying  according 
to  these  bases,  we  obtain  :  denominators  the  same  and  fraction  in  the 
minuend  the  greater,  denominators  the  same  and  fraction  in  the  min- 
uend the  smaller ;  denominators  different  and  fraction  in  the  minu- 
end the  greater,  denominators  different  and  fraction  in  the  minuend 
the  smaller. 

Thus  :   8|  -  5  J,  8J  -  5| ;  8f  -  5|,  8§  -  5|. 

9.  Measurements.     The  teacher  should  have  in  mind 
the  steps  in  all  measurement  as  a  guide  to  skilful  presen- 
tation. 

To  measure  an  object,  assert  that  it  possesses  as  much 
of  a  quality  as  a  well-known  standard.  Abbreviate  the 
concept  by  a  name.  Thus,  the  act  requires  as  much 
time  as  the  rotation  of  the  earth  about  its  axis,  or  a 
day. 

If  the  object  possesses  as  much  of  the  quality  as  the 
standard,  as  much  more,  as  much  more,  and  so  on,  intro- 
duce number.  Thus,  the  act  requires  as  much  time  as 
two  rotations,  or  two  days. 

If  the  measurement  requires  a  large  number  of  repeti- 
tions of  the  standard,  select  a  larger  standard.  Thus,  the 
act  requires  as  much  time  as  the  revolution  of  the  earth 
about  the  sun,  or  a  year. 

If  the  object  possesses  less  of  the  quality  than  the 
standard,  select  a  smaller  standard.  Thus,  the  act  requires 
as  much  time  as  the  24th  of  a  day,  or  an  hour. 


§  10  LESSON  2.    MEASUREMENTS  9 

If  necessary,  use  two  or  more  standards.  Thus,  the  act 
requires  5  years  2  days  and  3  hours. 

ILL.  Length,  T.  "  How  long  is  that  line  on  the  board  ?  About 
so  long  (the  hands  are  held  apart).  How  far  from  your  desk  to  the 
door?  16  steps  (the  pupil  counts  his  steps). 

"  It  is  impossible  to  hold  the  hands  the  exact  distance  apart ;  steps 
are  not  all  of  the  same  length.  Here  is  a  rule  which  is  as  long  as  a 
certain  king's  foot ;  it  is  called  a  foot.  Mary  may  measure  the  line  ; 
it  is  2  feet  long. 

"  Here  is  a  rule  3  feet  long;  it  is  called  a  yard ;  it  is  more  convenient 
for  measuring  long  distances  than  a  foot  rule ;  it  is  used  in  meas- 
uring cloth.  John,  measure  the  long  line  ;  it  is  2  yards  long. 

"  With  the  foot  rule  Henry  may  measure  the  short  line ;  it  is  1  foot 
and  a  little  more.  To  measure  this  little  more,  what  must  we  have? 
Yes,  a  still  shorter  rule.  This  foot  rule  has  been  divided  into  12 
equal  parts,  and  each  part  is  called  an  inch.  Measure  the  line  again, 
Henry ;  it  is  1  foot  4  inches.  Who  will  give  me  the  table  ?  12  inches 
make  1  foot,  3  feet  make  1  yard." 

10.  Logical  Steps.     Preparatory  to  many  exercises,  the 
teacher  must  discover  the  steps  which  are  taken  in  com- 
mon practice  to  reach  the  desired  end. 

Perform  the  exercise  and  analyze  the  steps. 

ILL.  Multiplication.  Let  us  discover  the  logical  steps  in  the  mul- 
tiplication of  an  integer  by  a  number  of  two  orders. 

In  multiplying  264  by  24,  we  multiplied  264  by  4,  we  multiplied 
264  by  20,  and  added  the  results.  The  logical  steps  are  to  multiply 
by  the  number  in  units'  order,  to  multiply  by  the  number  in  tens' 
order  and  the  result  by  10,  and  to  add  the  products. 

11.  Exercises.     1.   Teach  pupils  to  find  the  problems  which  grow 
out  of  the  statement,  The  interest  of  $200  for  2  years  at  6%  is  824. 
2.   From  the  simple  interest  formulae,  /  =  P  x  T  x  R  and  A  •=  P  +  7, 
find  the  20  cases.    3.   Why  are  the  cases  API  to  find  T  and  R,  impos- 
sible?     4.  The  following  divisors  are  to  be  used  in  long  division,  71 
and  17.     (a)  Arrange  them  in  order  of  difficulty;  (6)  give  reasons 
for  your  arrangement.      6.   Teach    quart,   pint,    gallon,   in   liquid 
measure.      6.   State  the  logical  steps  in  the  addition  of  fractions. 


LESSON   3.     ONE-LINE  DIAGRAMS 

12.  One-Line  Diagrams.  To  get  clear  notions  of  the 
various  operations  upon  fractions  pupils  should  be  taught 
to  make  and  use  diagrams  for  themselves.  Diagrams 
made  by  teachers  do  little  good  to  pupils.  One-line  dia- 
grams are  based  upon  separating  a  line  into  a  number  of 
equal  parts  and  then  into  another  number  of  equal  parts 
in  such  a  way  as  to  show  a  common  measure. 

ILL.  T.  "  We  wish  to  separate  a  line  into  8  equal  parts  and  then 
into  6  equal  parts  so  as  to  show  a  common  measure. 

"  How  shall  we  proceed  ?  Separate  the  line  by  short  vertical  lines 
into  8  equal  parts. 


"Then  what?  The  least  number  which  exactly  contains  8  and  6 
is  24.  Separate  the  line  by  dots  into  24  equal  parts.  How  can  we  do 
this?  The  line  is  already  separated  into  8  equal  parts  and  we  can 
separate  each  small  portion  into  24  -j-  8  or  3  equal  parts.  Do  so. 


I  '  •  i  -  •  i  •  •  t  •  •  I  •  '  I  '  •  t  ' — I 

"Now  what?  We  want  to  separate  the  line  into  6  equal  parts. 
How  many  dots  to  a  part  ?  24  -*-  6  or  4.  Draw  a  short  vertical  line 
above  at  every  4th  dot. 


I I        .  i 


•  I  '  •  i 

"What  does  each  vertical  line  below  show?  £.  What  does  each 
dot  show?  ^?.  What  does  each  vertical  line  above  show?  J. 
What  is  a  common  measure  of  £  and  |?  -fa.  How  many  24ths  make 
1  8th?  3.  How  many  24ths  make  1  6th?  4." 

10 


§13  LESSON  3.    ONE-LINE  DIAGRAMS  11 

Designating  Parts.  Designate  a  required  portion  by 
placing  its  value  in  the  center  of  a  dotted  line  which  con- 
nects the  extremities  of  the  portion. 

ILL.  T.  "  Let  us  draw  a  diagram  to  represent  f  and  f .  How  shall 
we  proceed?  Divide  a  line  into  4  equal  parts  and  then  into  3  equal 
parts.  Mark  off  J  and  f  as  I  have  done." 


Uses.  Pupils  should  solve  examples  in  addition,  sub- 
traction, multiplication,  and  division  of  fractions  as  a 
preparation  for  solutions  by  rules. 

ILL.  1.  T.  "  You  have  just  made  a  diagram  to  represent  £  and  f . 
Find  the  value  of  f  +  f .  The  value  is  &  +  T8j  or  f  J. 

"  Find  the  value  of  £  -  f .     The  value  is  ^  -  T\  or  ^. 

u  Find  the  value  of  f  -4-  |.     The  value  is  8  12ths  -^  9  12ths  or  f ." 

ILL.  2.  T.  "Let  us  find  f  of  f.  What  must  we  do?  Represent!, 
divide  |  into  3  equal  parts,  and  take  2  parts. 


"  f  of  |  is  8  of  the  15  equal  parts  into  which  the  unit  is  divided  or 

i  of  |  is  A-" 

13.  Mechanical  Aids.  Pupils  should  be  taught  to  use 
mechanical  devices  especially  in  mensuration. 

Paper  Folding  and  Cutting.  Let  us  agree  that  the 
pupil  shall  do  the  work. 

ILL.     Triangles.     T.  "  What  is  the  sum  of  the  angles  of  a  triangle  ? 

"How  shall  we  proceed?  Let  us  cut  a  triangle  from  paper  and 
fold  in  such  a  way  as  to  bring  the  vertices  together  at  the  same  point. 
Fold  the  upper  vertex  over  \ipon  the  base  so  as  to  make  the  folded 
edge  parallel  to  the  base.  Fold  the  other  parts.  What  seems  to  be 


12 


LESSON   3.     MECHANICAL  AIDS 


§13 


the  sum  of  the  angles  of  a  triangle?  The  sum  of  the  angles  of  a 
triangle  seems  to  be  the  sum  of  the  angles  about  a  point  on  one  side 
of  a  straight  line  or  180°." 


ILL.  Parallelograms.  T.  "  We  want  a  rule  for  finding  the  area  of  a 
parallelogram.  We  know  how  to  find  the  area  of  a  rectangle.  Let 
us  find  how  the  area  of  a  parallelogram  compares  with  the  area  of  a 
rectangle.  How  shall  we  make  this  comparison?  Cut  a  parallelo- 
gram from  paper.  Starting  from  one  of  the  vertices  cut  off  a  right- 
angle  triangle.  Put  the  triangle  first  at  one  end  of  the  second  por- 
tion and  then  at  the  other.  What  seems  to  be  true  ?  The  area  of  a 
parallelogram  seems  to  be  the  same  as  the  area  of  a  rectangle  which 
has  the  same  base  and  altitude." 


Crude  Measurements.  Some  things  are  hard  to  measure, 
as  the  circumference  of  a  circle  or  the  surface  of  a  sphere. 
Pupils  should  be  asked  to  exercise  their  ingenuity  on  such 
measurements. 

ILL.  Circumference  of  a  Circle.  T.  "  Here  is  a  circle  on  the  board. 
How  shall  we  find  its  circumference?  Sometimes  the  blacksmith 
wants  to  cut  off  a  strip  of  steel  long  enough  to  make  the  tire  of  a 
wagon  wheel.  Do  you  know  how  he  finds  the  proper  length  ?  He 
has  a  rule  made  in  the  form  of  a  circle,  with  a  handle.  He  finds  how 
many  times  this  small  wheel  turns  around  in  moving  about  the  cir- 
cumference of  the  wagon  wheel. 


"  Can  we  use  such  a  wheel  to  measure  the  circumference  of  this 
circle  on  the  board  ?    What  shall  we  use?    I  have  a  piece  of  electric 


§  14  LESSON  3.    MECHANICAL  AIDS  13 

wire  here.  How  can  we  use  it  ?  John  may  lay  the  wire  on  the  cir- 
cumference and  then  measure  the  wire.  What  is  the  length  of  the 
circumference?  44  in. 

"  Let  us  see  if  we  can  find  an  easier  way.  Measure  the  diameter  of 
the  circle.  It  is  14  in.  Divide  44  in.  by  14  in.  The  answer  is  3f. 
What  seems  to  be  the  circumference  of  a  circle  ?  The  product  of  its 
diameter  by  3}." 

ILL.  Surface  of  a  Sphere.  T.  "  How  shall  we  measure  the  surface 
of  this  sphere? 


"  I  have  here  a  piece  of  waxed  cord  and  the  half  of  a  croquet 
ball  (a  hemisphere)  into  which  I  have  driven  two  tacks.  I  propose  to 
wrap  the  cord  about  one  tack  until  the  string  covers  the  curved 
surface,  and  then  about  the  other  tack  until  the  cord  covers  the  plane 
surface,  and  then  to  compare  the  lengths.  Henry  may  do  this.  The 
cord  about  the  curved  surface  is  twice  the  length  of  that  about  the 
plane  surface.  What  does  this  seem  to  show  ?  The  surface  of  a  hemi- 
sphere seems  to  be  twice  the  surface  of  a  circle  which  has  the  same 
diameter,  or  the  surface  of  a  sphere  seems  to  be  4  times  the  surface 
of  a  circle  which  has  the  same  diameter." 

14.  Exercises.  1.  Divide  a  line  into  4  equal  parts  and  then  into 
8  equal  parts.  2.  Represent  |  and  J.  3.  By  diagram  find:  (a)  f  +  f  ; 
(*)  $  —  1 5  (c)  $  -*- f  •  4-  By  diagram  find  |  of  f .  6.  By  paper  cut- 
ting find  the  relation  of  a  triangle  to  a  parallelogram  which  has  the 
same  base  and  altitude.  6.  Cut  a  triangle  from  paper  and  fold  as 
in  finding  the  sum  of  its  angles.  Show  that  the  area  of  a  triangle 
seems  to  be  the  area  of  a  rectangle  which  has  the  same  base  and  half 
the  same  altitude. 


LESSON  4.     PRINCIPLES  —  INDUCTION 

15.  Inductive  Method.      There  are  many  principles  in 
arithmetic  which  must  be  established  as  guides  to  methods 
of  procedure.     It  is  better  for  the  pupil  to  develop  these 
for   himself    than    to    study  them    ready  formulated   by 
another.     The  inductive  method  is  the  process  of  estab- 
lishing principles  by  experiment.     Five  canons  of  induc- 
tion are  discussed  in  logic,  but  only  the  canon  of  agree- 
ment will  be  considered  in  this  treatise.     See  logic  (J.  &  H., 
p.  215). 

16.  Canon  of  Agreement.     Whatever  is  true  of  several 
individuals  of  a  class  is  probably  true  .of   all   the   indi- 
viduals of  that  class.     Thus,  by  experiment  it  is  found  of 
several  examples  in   subtraction  that   adding   the   same 
number  to  both  minuend  and  subtrahend  does  not  affect 
the  remainder  ;  this  principle  is  probably  true  of  all  ex- 
amples in  subtraction.     When  a  proposition  is  established 
in  regard  to  individuals  by  experiment  alone,  the  inference 
in  regard  to  the  class  can  never  be  more  than  probable 
unless  every  individual  in  the  class  has  been  examined. 
The  greater  the  number  of   individuals  the  greater  the 
probability. 

Take  several  instances  in  which  the  phenomenon  occurs, 
examine  these  instances  for  a  common  circumstance,  and 
make  the  common  circumstance  the  basis  of  a  general- 
ization. 

ILL.  T.  "  We  wish  to  discover  a  rule  for  the  divisibility  of  a 
number  by  9. 

U 


§16  LESSON  4.        PRINCIPLES  — INDUCTION  15 

"  Let  us  examine  several  numbers  which  are  exactly  divisible  by 
9.  To  get  such  numbers  those  who  sit  in  the  first  row  may  multiply 
a  number  of  three  figures  by  9 ;  those  in  the  second  row  a  number 
of  4  figures ;  all  others,  a  number  of  5  figures.  Read  some  of  the 
products :  2034,  3861,  23895,  808884.  Do  you  find  anything  which 
these  numbers  have  in  common?  Get  the  sums  of  the  digits  —  9, 
18,  27,  36.  Now  do  you  find  anything  in  common?  The  sum  of  the 
digits  is  exactly  divisible  by  9.  What  seems  to  be  the  rule  for  the 
divisibility  of  a  number  by  9?  A  number  seems  to  be  divisible 
by  9  if  the  sum  of  its  digits  is  divisible  by  9.  Memorize  this  rule." 

Weakened  Form.  An  inference  from  a  single  instance 
can  have  little  weight  of  itself,  but  is  of  value  in  a  few 
cases  where  its  establishment  by  other  methods  is  too 
difficult  for  the  grade.  For  examples,  see  §  13.  It  plays 
an  important  part  also  in  the  complete  method  (§  20). 
Its  use  elsewhere  is  not  recommended.  Observe  the 
weakness  of  the  following  : 

ILL.  T.  "Find  by  diagram  f  of  f.  The  value  is  &  (§12). 
What  seems  to  be  a  rule  for  multiplying  fractions?  To  multiply 
fractions  multiply  the  numerators  for  a  new  numerator,  and  the 
denominators  for  a  new  denominator." 

Use  in  Arithmetic.  Except  as  a  part  of  the  complete 
method,  the  inductive  method  should  be  rarely  used  in 
arithmetic.  An  examination  of  enough  instances  to 
warrant  a  conclusion  is  long  and  laborious  ;  no  appeal  is 
made  to  the  intelligence  ;  an  experimental  method  is  not 
suited  to  an  exact  science  ;  and  at  the  best,  the  conclusion 
can  never  be  more  than  probable.  Used  by  itself,  it  is 
valuable  for  finding  rules  for  divisibility  in  the  lower 
grades,  and  for  rinding  a  few  rules  for  mensuration  in  the 
upper  grades,  because  in  both  cases  the  pupils  are  not 
sufficiently  mature  to  use  more  satisfactory  methods. 


16  LESSON  4.    PRINCIPLES  — DEDUCTION  §17 

17.  Deductive  Method.      The  deductive  method  is  the 
method  of  establishing  principles  by  giving  reasons  for  the 
steps.     Several  forms  of  deduction  are  discussed  in  logic 
but  only  the  form,  A  is  B,  B  is  C,  .  *.  A  is  C,  will  be  con- 
sidered in  this  treatise.     See  logic  (J.  &  H.,  p.  145). 

18.  A  is  B,  etc.      Whatever  is  true  of  a  term  is  true  of 
what  is  included  within  that  term  or  of  what  is  identical 
with  that  term.      The  argument  may  take  the  form  A  is 
B,  B  is  C,  . '.  A  is  C,  or  the  form  of  a  chain  of  such  argu- 
ments, A  is  B,  B  is  (7,  C  is  D,  D  is  E,  .-.  A  is  E.     Each 
premise  must  be  established  by  a  definition,  an  axiom,  or 
a  proposition  previously  proved. 

ILL.  Multiplying  both  numerator  and  denominator  of  a  fraction 
by  the  same  number  first  multiplies  the  fraction  by  a  number  and 
then  divides  the  result  by  the  same  number,  because  multiplying  the 
numerator  multiplies  a  fraction  and  multiplying  the  denominator 
divides  a  fraction  (A  is  E). 

Multiplying  a  fraction  by  a  number  and  dividing  the  result  by  the 
same  number  does  not  change  the  value  of  a  fraction  by  axiom  (B 
is  C). 

Therefore,  multiplying  both  numerator  and  denominator  of  a  frac- 
tion by  the  same  number  does  not  change  the  value  of  a  fraction  (A 
is  C). 

Make  some  predication  about  the  subject  of  the  required 
proposition  based  upon  a  definition,  an  axiom,  or  a  propo- 
sition already  proved.  This  gives  the  form,  A  is  B. 

Make  some  predication  about  the  predicate  of  the  last 
proposition  based  upon  a  definition,  an  axiom,  or  a  propo- 
sition already  proved.  This  gives  the  form,  B  is  0. 

Continue  as  before  until  a  serviceable  predicate  is  found, 
and  unite  it  with  the  subject  of  the  first  proposition.  This 
may  give  the  form,  A  is  E. 

Individual  Subject.     In  teaching,  it  is  better  to  use  an 


§19  LESSON  4.    PRINCIPLES  — DEDUCTION  17 

individual  than  a  general  term  because  the  process  is 
then  more  vivid.  Thus,  '  multiplying  the  numerator  and 
denominator  of  |  by  2  '  is  more  vivid  than  '  multiplying 
both  numerator  and  denominator  of  a  fraction. by  the  same 
number.'  ,  Whatever  is  proved  of  the  individual  in  this 
way  is  proved  of  the  whole  class  which  includes  that 
individual  because  the  principle  could  be  proved  of  every 
other  individual  of  the  class  in  the  same  way. 

ILL.  Multiplication  of  Decimals.  T.  "  We  are  going  to  learn  how 
to  multiply  a  decimal  by  a  decimal.  We  know  how  to  multiply  a 
decimal  by  an  integer  and  how  to  multiply  by  .1,  .01,  and  so  on. 

.24 

.6 

.144 

«  Take  .24  x  .6.  What  is  to  multiply  by  .6?  To  multiply  by  6  and 
to  multiply  the  result  by  .1  because  .6  is  6*x  .1. 

"You  may  all  multiply  by  6;  the  answer  is  1.44;  to  multiply  a 
decimal  by  an  integer  multiply  as  in  integers  and  point  off  as  many 
decimal  places  in  the  product  as  there  are  decimal  places  in  the 
multiplicand.  Multiply  the  result  by  .1;  the  answer  is.lx44;  to 
multiply  by  .1,  .01,  .001,  and  so  on,  move  the  decimal  point  as  many 
places  to  the  left  as  there  are  decimal  places  in  the  multiplier. 

"  What  is  to  multiply  .24  by  .6  or  to  multiply  a  decimal  by  a  deci- 
mal ?  Multiply  as  in  integers  and  point  off  as  many  decimal  places 
in  the  product  as  there  are  decimal  places  in  both  multiplicand  and 
multiplier." 

19.  Exercises.  1.  Using  the  method  of  agreement,  help  pupils  to 
find  a  rule  for  the  divisibility  of  a  number  by  4.  2.  Determine 
whether  the  formula,  x'2  +  x  +  41  =  a  prime,  is  true  for  all  integral 
values  of  x.  Suggestion.  For  x  =  0,  1,  2,  3,  4,  5,  6,  the  values  are  41, 
43,  47,  53,  61,  71,  83,  respectively.  3.  Using  the  form,  A  is  B,  B  is 
C,  and  so  on,  assist  pupils  to  find  how  to  multiply  integers  by  a 
number  of  two  orders.  4.  Why  is  the  deductive  method  better  than 
the  inductive  method  for  establishing  the  principles  of  arithmetic? 


LESSON   5.     PRINCIPLES  —  COMPLETE  METHOD 

20.  The  Complete  Method.  The  race  has  established 
many  important  principles  in  mathematics  by  inferring 
some  property  from  an  examination  of  one  or  more  indi- 
viduals and  by  discovering  why  the  property  must  be  true 
of  all  the  individuals  of  the  class.  Thus,  the  race  found 
by  measurement  that  of  a  right-angle  triangle  whose  legs 
are  3  in.  and  4  in.  the  hypotenuse  is  5  in.  From  this  and 
other  measurements  they  inferred  that  the  square  of  the 
hypotenuse  of  every  right-angle  triangle  must  be  the  sum 
of  the  squares  of  the  other  two  sides.  By  the  deductive 
methods  of  geometry  they  proved  that  their  surmise  was 
correct.  The  complete  method  consists  in  finding  a  prin- 
ciple by  experiment  (induction)  and  in  proving  it  by  rea- 
sons (deduction).  It  is  the  method  of  the  discoverer. 
See  logic  (J.  &  H.  p.  249). 

Find  by  experiment  a  proposition  which  is  true  of  one 
or  more  individuals  of  a  class  and  discover  from  defini- 
tions, axioms,  or  propositions  already  proved,  why  it  must 
be  true  of  the  whole  class. 

ILL.  Mult.  Terms  of  Fraction.  T.  "  We  wish  to  discover  the  effect 
on  a  fraction  of  multiplying  its  numerator,  multiplying  its  denomina- 
tor, and  multiplying  both  numerator  and  denominator  by  the  same 
number. 

"  You  may  draw  a  diagram  like  mine  showing  f ,  f ,  and  f . 

-          *         .  * 


18 


§20          LESSON  5.    PRINCIPLES  — COMPLETE  METHOD  19 

"  Let  us  discover  the  effect  of  multiplying  the  numerator.  Take  f  and 
multiply  the  numerator  by  2 ;  the  result  is  f .  From  the  diagram,  tell 
me  what  has  been  done  to  the  fraction  ;  it  has  been  multiplied  by  2. 
What  seems  to  be  the  effect  of  multiplying  the  numerator?  To  mul- 
tiply the  fraction.  Who  can  tell  me  why?  Multiplying  the  numera- 
tor multiplies  the  number  of  equal  parts  taken  without  affecting  the 
size  of  the  parts.  Write  the  rule,  multiplying  the  numerator  multi- 
plies the  fraction. 

"Let  us  discover  the  effect  of  multiplying  the  denominator.  Take  f. 
and  multiply  the  denominator  by  2;  the  result  is  |.  From  the  dia- 
gram, tell  me  what  has  been  done  to  the  fraction  ;  it  has  been  divided 
by  2.  What  seems  to  be  the  effect  of  multiplying  the  denominator? 
To  divide  the  fraction.  Who  can  tell  'me  why  ?  What  was  the  size 
of  one  of  the  equal  parts  before  we  multiplied  the  denominator?  A 
fourth.  After  we  multiplied  the  denominator?  An  eighth.  What 
did  we  do  to  the  size  of  one  of  the  equal  parts  ?  Divided  it  by  2. 
Now  who  can  tell  why?  Multiplying  the  denominator  divides  the 
size  of  the  equal  parts  without  affecting  the  number  of  parts  taken. 
Write  the  rule,  multiplying  the  denominator  divides  the  fraction. 

"  Let  us  discover  the  effect  of  multiplying  both  terms  by  the  same 
number.  Take  \  and  multiply  both  terms  by  2 ;  the  result  is  f.  From 
the  diagram,  tell  me  what  has  been  done  to  the  fraction  ;  its  value  has 
not  been  changed.  What  seems  to  be  the  effect  on  a  fraction  of  multi- 
plying both  terms  by  the  same  number?  It  does  not  change  the  value 
of  the  fraction.  Who  can  tell  me  why?  When  we  multiplied  the 
numerator  by  2  what  did  we  do  to  the  fraction?  We  multiplied  the 
fraction.  When  we  multiplied  the  denominator  of  the  result  by  2 
what  did  we  do  to  the  result?  We  divided  the  result  by  2.  If  we 
multiply  a  fraction  by  2  and  divide  the  result  by  2  what  do  we  do  to 
the  fraction?  Write  the  rule,  multiplying  both  numerator  and  de- 
nominator of  a  fraction  by  the  same  number  does  not  change  the 
value  of  the  fraction." 

Brief  Rules.  After  several  rules  of  kindred  nature  have 
been  established  it  is  sometimes  possible  to  make  a  brief 
rule  which  comprehends  them  all.  A  valuable  illustra- 


20  LESSON  5.    PRINCIPLES  — COMPLETE  METHOD          §20 

tion  is  found  in  the  principles  of  multiplying  just  devel- 
oped, together  with  the  principles  of  division  that  may  be 
developed  in  a  similar  manner. 

ILL.  T.  "Multiplying  the  numerator  multiplies  the  fraction,  di- 
viding the  numerator  divides  the  fraction.  Who  can  express  the  two 
rules  by  one  ?  In  the  case  of  multiplying  and  dividing,  doing  either 
thing  to  the  numerator  does  the  same  thing  to  the  fraction.  Let  us 
put  it  shorter.  Doing  a  thing  to  the  numerator  does  the  same  thing  to  the 
fraction. 

"Multiplying  the  denominator  divides  the  fraction, dividing  the  de- 
nominator multiplies  the  fraction.  Give  me  a  short  rule  for  these 
two.  Doing  a  thing  to  the  denominator  does  the  opposite  thing  to  the  frac- 
tion. 

"  Multiplying  both  numerator  and  denominator  by  the  same  number 
does  not  change  the  value  of  the  fraction,  dividing  both  numerator 
and  denominator  by  the  same  number  does  not  change  the  value  of  the 
fraction.  Give  me  a  short  rule  for  these  two.  Doing  the  same  thing  to 
both  numerator  and  denominator  does  not  change  the  value  of  the  fraction."1' 

Use  in  Arithmetic.  The  complete  method  should  be 
used  in  developing  nearly  all  the  principles  of  arithmetic 
because  it  is  the  method  of  the  discoverer.  It  discovers 
something  by  trial  that  may  be  true  and  proves  by  reasons 
that  it  is  true. 

ILL.  Multiplying  Fractions.  T.  "To-day  we  are  going  to  learn 
how  to  multiply  a  fraction  by  a  fraction.  Take  f  of  f  and  find  the 
result  by  diagram;  ^5.  (See  diagram, p.  11.) 

"  How  can  8  be  obtained  from  the  numerators  2  and  4  ?  How  can 
15  be  obtained  from  the  denominators  3  and  5?  What  seems  to  be 
the  rule  for  multiplying  fractions?  Multiply  the  numerators  for  a 
new  numerator  and  the  denominators  for  a  new  denominator. 

"  Let  us  find  why  this  rule  is  true.  What  does  f  of  f  mean?  That 
£  is  to  be  divided  by  3  and  the  result  multiplied  by  2  because  the  de- 
nominator shows  into  how  many  equal  parts  a  thing  is  to  be  divided, 
and  the  numerator  shows  how  many  equal  parts  are  taken. 


§21          LESSON  5.    PRINCIPLES  — COMPLETE   METHOD          21 

"  Divide  -  by  3 ;     ,   because  multiplying  the   denominator 

5  5x3 

4x2 
divides  the  fraction.  Multiply  the  result  by  2  ;  — —  ,  because  rnul- 

5x3 
tiplying  the  numerator  multiplies  the  fraction. 

9        A      4x2 

"  Write,  -  of  -  = Now  we  see  why  the  rule  is  true. 

3        55x3 

u  To  multiply  fractions,  multiply  the  numerators  for  a  new  numera- 
tor, and  the  denominators  for  a  new  denominator.  Write  the  rule." 

ILL.  Dividing  Fractions.  T.  "  We  want  a  rule  for  dividing  a 
fraction  by  a  fraction. 

"  By  diagram,  divide  f  by  f ;  the  result  is  f .     (See  diagram,  p.  11.) 

"How  could  we  get  f?  By  inverting  the  divisor  and  proceeding 
as  in  multiplication ;  f  x  $  =  f .  What  may  possibly  be  a  rule  for 
dividing  a  fraction  by  a  fraction  ?  Invert  the  divisor  and  proceed  as 
in  multiplication. 

"  Let  us  see  if  we  can  find  reasons  for  this  rule.  The  other  day  we 
learned  how  to  divide  1  by  a  fraction ;  invert  the  divisor.  What  is 
1  -H  |?  f  If  '1  divided  byf '  is  f,  whatis|of  '1  divided  by  £'?  f  of  f. 
The  rule  must  be  true.  To  divide  a  fraction  by  a  fraction  invert  the 
divisor  and  proceed  as  in  multiplication.  Write  the  rule." 

21.  Exercises.  1.  Using  the  diagram,  p.  18,  teach  pupils  by  the 
complete  method  to  discover  the  effect  on  a  fraction :  (a)  of  dividing 
the  numerator;  (ft)  of  dividing  the  denominator;  (c)  of  dividing 
both  numerator  and  denominator  by  the  same  number.  2.  Teach 
pupils  by  the  complete  method  how  to  divide  1  by  a  fraction. 
3.  Why  is  the  complete  method  better  for  No.  1 ;  (a)  than  the  induc- 
tive method  alone  ?  (6)  than  the  deductive  method  alone  ?  4.  What 
is  the  advantage  of  the  brief  rules  suggested  for  the  principles  in 
fractions  ? 


LESSON  6.     SIMPLE  PROBLEMS  —  BY  EXPERIMENT 

22.  Kinds  of  Problems.      An  exercise  involving  num- 
ber must  state  the  operations  directly  or  indirectly;  this 
gives  rise  to  examples  and  problems.     A  problem  must 
involve  one  operation  or  more  than  one  ;  this  gives  rise  to 
simple  problems  and  complex  problems.     A  simple  prob- 
lem must   involve   addition,   subtraction,   multiplication, 
the  first  case  in  division  known  as  quotition,  or  the  second 
case  in  division  known  as  partition. 

ILL.  Example.   Multiply  3 1  by  5. 

ILL.  Simple  Problem.   At  3  ^  each  what  is  the  cost  of  5  apples  ? 

ILL.  Complex  Problem.    If  2  apples  cost  6  f,  how  much  do  5  apples 
cost? 

23.  Importance  of  Simple  Problems.      The  solution  of 
simple  problems  is  the  basis  of  the  solution  of  all  prob- 
lems, because  a  complex  problem  can  be  separated  into  a 
chain  of  simple  problems,  and  can  be  solved  by  solving 
each  simple  problem  in  succession. 

24.  Stages  of   Progress.      In  the  advancement  of  the 
race  there  have  been  three  stages ;  the  stage  of  obtaining 
results  by  counting,  the  stage  of  obtaining  results  chiefly 
by  addition,  the  stage  of    obtaining  results  by  the  most 
fitting  operation.     These  three  stages  should  be  observed 
in  the  advancement  of  the  child. 

ILL.  At  3  f  each  what  is  the  cost  of  5  apples  ? 


OOO   OOO  OOO  OOO  OOO 

Counting  Stage.     Here  are  5  apples,  and  3  cents  for  each  apple. 
They  cost  1,  2,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  14,  15  cents. 
Addition  Stage.     They  cost  3,  6,  9,  12,  15  cents. 
Advance  Stage.     They  cost  5  times  3  cents,  or  15  cents. 


§25      LESSON  6.    SIMPLE  PROBLEMS— BY  EXPERIMENT      23 

25.  Counting  Stage.  As  soon  as  pupils  can  count  they 
have  at  command  a  means  for  solving  the  problems  of 
their  daily  experience.  They  can  represent  the  terms  by 
objects  and  find  the  results  by  counting.  This  work  ex- 
ercises their  ingenuity,  gives  them  a  feeling  of  power,  and 
lays  the  foundation  for  all  subsequent  work. 

ILL.  Simple  Prob.  in  Add.  On  each  desk  there  is  a  bundle  of 
sticks. 

T.  "If  John  has  3  apples  and  Joseph  has  2  apples,  how  many 
apples  have  both  ?  Use  the  sticks  for  apples  and  find  out.  Mary  may 
explain." 

M.  "  Here  are  John's  apples,  1,  2,  3  ;  here  are  Joseph's,  1,  2  ;  both 
have  1,  2,  3,  4,  5  apples." 

ILL.  Simple  Prob.  in  Sub.  T.  "Susan  had  6^  and  spent  2?. 
How  many  cents  did  she  have  left?  Use  the  sticks  for  cents  and 
find  out.  Henry  may  explain."  • 

<p  ©  ©  ©  ©  ® 

H.  "  Here  are  the  6 1  she  had,  1,  2,  3,  4,  5,  6 ;  she  spent  1,  2  cents  : 
she  had  left  1,  2,  3,  4  cents." 

ILL.  Simple  Prob.  in  Mult.  On  each  desk  there  are  sticks  and 
pieces  of  paper.  T.  "  A  boy  found  3  nests ;  there  were  2  eggs  in 
each  nest.  How  many  eggs  did  he  find?  Use  the  pieces  of  paper 
for  nests  and  the  sticks  for  eggs.  Susan  may  explain." 


S.  "  Here  are  the  three  nests,  1,  2,  3 ;  here  are  the  2  eggs  in -each 
nest.  He  found  1,  2,  3,  4,  5,  6  eggs." 

ILL.  Simple  Prob.  in  Quotition.  T.  "  A  boy  found  8  eggs  ;  there 
were  2  eggs  in  each  nest.  How  many  nests  were  there?  Use  the 
pieces  of  paper  for  nests  and  the  sticks  for  eggs.  Joseph  may 
explain." 


24      LESSON  6.    SIMPLE  PROBLEMS  — BY  EXPERIMENT      §26 
OO  OO  00  OO 


J,  "  Here  are  the  8  eggs,  1,  2,  3,  4,  5,  6,  7,  8 ;  these  2  eggs  need  a 
nest;  these  2  need  a  nest;  these  2  need  a  nest;  these  2  need  a  nest; 
the  eggs  are  all  used  ;  there  are  1,  2,  3,  4  nests." 

ILL.  Simple  Prob.  in  Partition.  T.  "  A  girl  had  6  oranges  and  2 
plates ;  she  put  the  same  number  of  oranges  on  each  plate.  How 
many  oranges  did  she  put  on  each  plate  ?  Peter  may  explain." 


P.  "Here  are  the  6  oranges  and  the  2  plates;  I  put  an  orange  on 
each  plate ;  I  put  another  orange  on  each  plate  ;  I  put  another  orange 
on  each  plate ;  the  oranges  are  all  placed  ;  there  are  1,  2,  3  oranges  on 
each  plate." 

26.  Addition  Stage.  Full  form.  As  soon  as  pupils  can 
add  they  have  a  still  better  means  for  solving  the  problems 
of  their  daily  experience.  They  can  represent  the  terms 
by  objects  and  find  the  results  by  addition. 

ILL.     S.  P.  in  Add.     See  §  25.     Both  have  3,  5  apples. 

ILL.     S.  P.  in  Mult.     See  §  25.     He  found  2,  4,  6  eggs. 

ILL.  S.  P.  in  Quot.  See  §  25.  Here  are  2,  4,  6,  8  eggs ;  they  need 
1,  2,  3,  4  nests. 

Short  Form.  In  solving  such  problems  as  are  suggested 
on  fruit  stands,  pupils  may  represent  groups  by  objects 
instead  of  individuals. 

ILL.  T.  "  At  4  for  5  ^  what  is  the  cost  of  12  oranges  ?  Make  a 
mark  for  each  group  of  4  oranges.  What  else  will  each  mark  repre- 
sent? A  group  of  5  cents.  Point  to  each  mark  and  add  by  5's.  Ann 
may  explain." 

A.  "  There  are  4,  8,  12  oranges ;  they  cost  5,  10,  15  cents." 


§27      LESSON   6.    SIMPLE  PROBLEMS  — BY  EXPERIMENT      25 

ILL.  T.  "At  3  for  5^  how  many  oranges  can  be  bought  for  25^? 
Shall  we  make  a  mark  for  each  group  of  3  oranges  or  for  each  group 
of  5  ^  ?  For  each  group  of  5  0,  because  the  number  of  cents  in  all  is 
given.  William  may  explain." 

I      I      I     I      I 

W.  "There  are  5,  10,  15,  20,  25  cents;  they  buy  3,  6,  9,  12,  15 
oranges." 

27.  Advance  Stage.     In  the  advance  stage,  pupils  select 
the  operation  best  suited  to  the  problem,  dispensing  with 
objects  except  in  cases  of  perplexity.     The  impulse  to  use 
an  operation  conies  as  soon  as  the  situation  is  grasped. 
If  the  situation  can  be  imaged  without  objects  no  objects 
are  necessary.     If  the  situation  cannot  be  imaged  objects 
are  imperative. 

ILL.  Problems  of  Paragraph  25.  They  both  have  3  apples  plus  2 
apples  or  5  apples.  She  had  left  6  $  minus  2 1  or  4  0.  The  boy  found 
3  times  2  eggs  or  6  eggs.  The  number  of  nests  is  the  number  of  times 
2  eggs  is  contained  in  8  eggs  or  4.  On  1  plate  the  girl  put  5  of  6 
oranges  or  3  oranges. 

28.  Exercises.     1.   State   an   example   involving  more   than  one 
operation.     2.    State  a  complex  problem  involving  three  operations. 
3.   Separate  the  complex  problem  into  its  simple  problems.     4.   State 
a  simple  problem  in  multiplication  and  solve  it  experimentally  by 
counting  objects.     6.   Solve  the  last  problem  experimentally  by  addi- 
tion.    6.    Solve  the  last  problem  experimentally  by  the  operation  in- 
volved.   7.   State  a  fruit  stand  problem  and  solve  it  experimentally 
by  representing  the  groups  by  objects. 


LESSON   7.     SIMPLE  PROBLEMS  —  BY  REASONS 

29.  Nature  of  Reasons.     Reasons  occur  in  pairs — the 
one  to  state  a  law  and  the  other  to  state  that  the  law 
applies.      Thus,    Smith   is   mortal   because   all   men   are 
mortal,  and  because  he  is  a  man.     In  the  last  lesson,  simple 
problems  are  solved  without  statement  of  reasons.     The 
situation  is  grasped  either  with  or  without  the  use  of 
objects  and  the  necessary  operation  is  applied.      Thus, 
Mary  has  2^  and  wants  to  spend  6^,  how  many  cents  does 
she  lack  ?     She  lacks  60  minus  2^  or  4^. 

Let  us  find  the  reasons.  We  subtract  2^  from  6^. 
Why  do  we  subtract  ?  Because  what  she  lacks  is  what 
she  must  spend  minus  what  she  has.  Why  do  we  perform 
the  operation  on  6^  and  2^?  Because  what  she  must 
spend  is  6^  and  what  she  has  is  2^. 

30.  Complete  Analysis.      The  complete  analysis  states 
both  reasons,  why  the  operation  is  selected  (the  law),  and 
why  the  terms  are  selected  (why  the  law  applies). 

ILL.  S.  P.  Add.  If  the  cost  is  Qp  and  the  gain  is  2?,  what  is  the 
selling  price? 

Since  the  selling  price  is  the  cost  plus  the  gain,  and  since  the  cost 
is  6^  and  the  gain  is  2f,  the  selling  price  is  6f*  plus  If  or  8^. 

ILL.  S.  P.  Sub.  If  the  selling  price  is  10  ?  and  the  gain  is  3  ^, 
•what  is  the  cost  ? 

Since  the  cost  is  the  selling  price  minus  the  gain,  and  since  the  sell- 
ing price  is  10^  and  the  gain  is  3^,  the  cost  is  10  f-  minus  3j«  or  7j*. 

ILL.  S.  P.  Mult.  If  the  cost  of  1  apple  is  3^,  what  is  the  cost  of 
5  apples  ? 

26 


§31  LESSON  7.    SIMPLE  PROBLEMS— BY  REASONS  27 

Since  the  cost  of  5  apples  is  5  times  the  cost  of  1  apple,  and  since 
the  cost  of  1  apple  is  3  /',  the  cost  of  5  apples  is  5  times  3  ^  or  15  f. 

ILL.  S.  P.  Quot.  If  the  cost  of  1  apple  is  3  0,  how  many  apples 
can  be  bought  for  15  ^  ? 

Since  the  number  of  apples  for  15^  is  the  number  of  times  the 
cost  of  1  apple  is  contained  in  15  ^,  and  since  the  cost  of  1  apple  is  3  ^, 
the  number  of  apples  for  15  ?  is  the  number  of  times  3  ^  is  contained 
in  15  ?  or  5. 

ILL.  S.  P.  Part.  If  the  cost  of  5  apples  is  15  j*,  what  is  the  cost  of 
1  apple? 

Since  the  cost  of  1  apple  is  £  the  cost  of  5  apples,  and  since  the 
cost  of  5  apples  is  15^,  the  cost  of  1  apple  is  ^  of  15  ^  or  3^. 

31.  Major  Analysis.     Usually  only  one  of  a  pair  of  rea- 
sons is  stated.     Thus,  Smith  is  mortal  because  all  men  are 
mortal,  or  Smith  is  mortal  because  he  is  a  man.      The 
major  analysis  retains  the  reason  for  the  selection  of  the 
operation. 

ILL.  Add.  Since  the  selling  price  is  the  cost  plus  the  gain,  the 
selling  price  is  6  ^  plus  2  ^  or  8  ^. 

ILL.  Sub.  Since  the  cost  is  the  selling  price  minus  the  gain,  the 
cost  is  10  ^  minus  3  f  or  7  ?. 

ILL.  Mult.  Since  the  cost  of  5  apples  is  5  times  the  cost  of  1 
apple,  the  cost  of  5  apples  is  5  times  3 1  or  15  t. 

ILL.  Quot.  Since  the  number  of  apples  for  15^  is  the  number  of 
times  the  cost  of  1  apple  is  contained  in  15  ?,  the  number  of  apples 
for  15^  is  the  number  of  times  3 1  is  contained  in  15^  or  5. 

ILL.  Part.  Since  the  cost  of  1  apple  is  £  the  cost  of  5  apples,  the 
cost  of  1  apple  is  \  of  15  J*  or  3^. 

32.  Minor  Analysis.     The  minor  analysis,  often  called 
the  model  analysis,  retains  the  reason  for  the  selection  of 
the  terms. 

ILL.  Add.  Since  the  cost  is  6  ^  and  the  gain  is  2  ^,  the  selling 
price  is  6  ^  plus  2  ^  or  8  ^. 

ILL.  Sub.  Since  the  selling  price  is  10  ^  and  the  gain  is  3  t,  the 
cost  is  10  ^  minus  3  ^  or  7 1. 


28  LESSON  7.    SIMPLE  PROBLEMS— BY  REASONS  §33 

ILL.     Mult.     Since  the  cost  of  1  apple  is  3  ^,  the  cost  of  5  apples  is 

5  times  3^  or  15^. 

ILL.  Qwo<.  Since  the  cost  of  1  apple  is  3  /,  the  number  of  apples 
for  15^  is  the  number  of  times  3f*  is  contained  in  15  fi  or  5. 

ILL.  Part.  Since  the  cost  of  5  apples  is  15  p,  the  cost  of  1  apple 
is  $  of  15 1  or  3  ?. 

33.  Use  in  Arithmetic.     The  solution  of  a  simple  prob- 
lem consists  in  grasping  the  situation  and  in  applying  the 
necessary  operation.     The  office  of  analysis  is  to  provide 
forms  of  expression  after  the  plan  has  been  formed.     Thus, 
we  discover  that  Mary  lacks  6^  minus  2^  as  soon  as  we 
realize  that  she  has  2  ^  and  wants  to  spend  6  /.     Some 
form  of  the  reason  may  be  in  the  mind  subconsciously,  but 
its  statement  comes  after  the  discovery  of  how  to  proceed. 
Hence,  it  is  a  mistake  to  teach  forms  of  analysis  as  aids  to 
the  solution  of  simple  problems.     The  true  use  of  analysis 
is  to  help  pupils  to  give  clear  forms  of  explanation  after 
the  solution  is  understood. 

The  model  analysis  seems  to  be  positively  harmful. 
Since  it  gives  the  reason  for  the  selection  of  the  terms 
without  indicating  the  operation,  it  draws  attention  away 
from  the  situation  and  fixes  it  upon  a  form  of  words. 

ILL.  Form  of  Mult.  Since  1  apple  costs  6^,  2  apples  cost  2 
times  6  ^. 

Misled  by  similarity  of  statement,  the  pupil  who  looks  for  aid  to 
model  analysis  instead  of  to  situations,  may  be  expected  to  say, 
"  Since  1  man  requires  6  days  for  a  work,  2  men  require  2  times  6 
days  or  12  days.  Since  a  dog  standing  on  1  leg  weighs  6  lb.,  stand- 
ing on  2  legs  he  weighs  2  times  6  lb.  or  12  lb.  Since  1  boy  gets  up  at 

6  o'clock,  2  boys  get  up  at  2  times  6  o'clock  or  12  o'clock." 

34.  Teaching.     After  pupils  reach  the  reason  stage,  it 
is  recommended  that  they   be  required   usually  to   give 
answers  without  any  form  of  explanation,  that  they  be 


§35  LESSON  7.    SIMPLE  PROBLEMS  — BY  REASONS  29 

asked  occasionally  for  an  explanation,  and  that  in  case  of 
failure  they  be  directed  to  study  the  situation. 

The  Usual.  The  pupil  speaks  or  writes  the  answer 
only;  all  thought  processes  are  unspoken  and  unwritten. 

ILL.  T.  "  If  the  selling  price  is  8  0  and  the  loss  is  2  ?,  what  is  the 
cost?"  P.  "10?." 

T.  "  If  a  boy  requires  1  hr.  to  walk  2  mi.,  how  many  hours  will  he 
require  to  walk  6  mi.  ?  "  P.  l<  3  hr. " 

The  Occasional.  The  pupil  states  the  answer,  and,  when 
asked  for  an  explanation,  gives  the  conclusion  of  the  com- 
plete analysis.  If  required,  he  states  also  the  reason  for 
the  selection  of  the  operation. 

ILL.  T.  "If  aboy  earns  $3  a  week,  in  how  many  weeks  will  he  earn 
$12?"  P.  "4."  T.  "Explain."  P.  «  He  will  earn  $12  in  as  many 
weeks  as  the  times  $3  is  contained  in  $12  or  4."  T.  "Why?"  P. 
"  Because  for  each  time  he  earns  $  3  there  will  be  1  week." 

In  Case  of  Failure.  The  pupil  is  directed  to  study  the 
situation  either  with  or  without  objects. 

ILL.  T.  "  How  many  barrels  are  required  for  12  bu.  of  apples  if  1 
bbl.  holds  3  bu.  ?  "  P.  "  I  cannot  tell.  "  T.  "  Take  bits  of  paper  for 
barrels  and  sticks  for  bushels  and  find  out."  P.  "  4  bbl."  T. 
"  Why  ?  "  P.  "  For  each  3  bu.  there  must  be  1  bbl.  There  must  be  as 
many  barrels  as  the  times  3  bu.  is  contained  in  12  bu.  or  4." 

T.  "  If  a  sum  amounts  to  $  1.06  in  1  yr.  at  simple  interest,  how  much 
will  it  amount  to  in  2  yr.  ?  "  P.  "  $2.12."  T.  "  State  the  parts  which 
make  up  $1.06."  P.  "The  principal  and  the  interest  for  1  yr."  T. 
"  Should  both  these  parts  be  multiplied  by  2  ?  Try  again."  P. 
"  The  amount  for  2  yr.  cannot  be  found." 

35.  Exercises.  1.  State  a  simple  problem  not  stated  in  the  lesson 
of  each  of  the  five  types  and  give  the  complete  analysis.  2.  State  the 
major  analysis  of  each.  3.  State  the  minor  analysis  of  each.  4. 
Help  a  pupil  who  fails  on,  If  2  men  require  6  da.  for  a  work,  how  many 
days  does  1  man  require  ?  5.  Why  should  pupils  be  required  usually 
to  give  answers  to  simple  problems  without  explanations? 


LESSON   8.      COMPLEX  PROBLEMS  —  BY  ARITHMETIC 

36.  Into   Simple   Problems.     A  complex  problem  is  a 
problem  that  involves  more  than  one  operation  (§  22).     It 
can  be  separated  into  a  chain  of  simple  problems  such  that 
the  answer  to  one  becomes  a  given  term  in  a  second,  the 
answer  to  a  second  becomes  a  given  term  in  a  third,  and 
so  on. 

To  solve  a  complex  problem  arithmetically,  state  a 
simple  problem  and  its  answer,  state  a  second  simple 
problem  and  its  answer,  and  so  on  until  the  answer  to  the 
last  simple  problem  is  the  answer  to  the  complex  problem. 

ILL.  A  boy  bought  10  Ib.  of  pop  corn  at  12  ^  a  pound  and  sold  it  at 
25^  a  pound.  What  was  his  entire  gain? 

If  the  cost  of  1  Ib.  was  12  f>  and  the  selling  price  25  ^,  what  was  the 
gain  on  1  Ib.  ?  13  p.  If  the  gain  on  1  Ib.  was  13  ^,  what  was  the  gain 
onlOlb.?  $1.30. 

37.  Finding  Component  Problems.     The  chief  work  in  the 
solution  of  a  complex  problem  is  to  find  the  component 
problems.     It  must  be  done  by  a  study  of  the  situation. 
There  are  two  principal  aids. 

Analytic  Aid.  The  value  of  the  term  required  in  the 
complex  problem  can  be  found  from  one  or  more  terms  by 
a  single  operation.  Thus,  the  entire  gain  can  be  found 
from  the  gain  on  1  Ib.  and  the  number  of  pounds. 
The  value  of  each  unknown  term  thus  used  can  be  found 
from  one  or  more  terms  by  a  single  operation,  and  so  on. 
Thus,  the  gain  on  1  Ib.  can  be  found  from  the  cost  of  1  Ib. 
and  the  selling  price  of  1  Ib.  As  soon  as  the  values  of 

30 


§37    LESSON  8.    COMPLEX  PROBLEMS  -BY  ARITHMETIC     31 

all  the  terms  are  known  the  simple  problems  can  be 
formed. 

Consider  from  what  terms  the  required  term  can  be 
found  by  a  single  operation,  consider  from  what  terms  each 
unknown  term  thus  used  can  be  found  by  a  single  opera- 
tion, and  so  on.  State  the  simple  problems. 

Make  a  diagram  by  joining  each  term  to  its  dependent 
terms. 

i-  •         1  1K  /Cost  lib.,  12? 
Gain  on  1  lb.<  ' 

\S.P.  1  Ib,  25  j* 


The  above  may  be  read,  The  entire  gain  can  be  found  from  the 
gain  on  1  Ib.  and  the  no.  of  pounds;  the  gain  on  1  Ib.  can  be  found 
from  the  cost  of  1  Ib.  and  the  selling  price  of  1  Ib. 

ILL.  1.  T.  "If  the  cost  of  3  apples  is  6^,  what  is  the  cost  of  5 
apples  ? 

"  From  what  can  we  find  the  cost  of  5  apples?  From  the  cost  of  1 
apple.  From  what  can  we  find  the  cost  of  1  apple  ?  From  the  cost  of 
3  apples. 

Cost  5  ap.  <  Cost  of  1  ap.  <  Cost  of  3  ap.,  6  ^. 

"  Solve.  If  the  cost  of  3  apples  is  6  ^,  what  is  the  cost  of  1  apple  ? 
2  p.  If  the  cost  of  1  apple  is  2  p,  what  is  the  cost  of  5  apples  ?  10  ^." 

ILL.  2.     T.  "  If  A  requires  2  da.  for  a  work  and  B  requires  3  da., 
how  many  days  do  both  require  ? 

"  From  what  can  we  find  the  no.  of  days  both  require  ?  From  the 
part  both  can  do  in  1  da.  From  what  can  we  find  the  part  both'  can 
do  in  1  da.  ?  From  the  part  A  can  do  in  1  da.  and  the  part  B  can  do 
in  1  da.  (and  so  on). 

/Part  A  1  da.  <  Days  A,  2 
Days  both  <  Part  both  1  da.<  _  J     „ 

\Part  B  1  da.  <  Days  B.  3 

"  Solve.  If  A  requires  2  da.,  what  part  can  he  do  in  1  da.  ?  \,  If 
B  requires  3  da.,  what  part  can  he  do  in  1  da.  ?  \.  If  A  can  do  \  of  it 
in  1  da.  and  B  can  do  \  of  it  in  1  da.,  what  part  can  they  both  do  in  1 
da.?  \.  If  both  can  do  |  of  it  in  1  da.,  in  how  many  days  can  they  do 
fofit?  l£da." 


32     LESSON  8.    COMPLEX  PROBLEMS-BY  ARITHMETIC    §38 

Grraphic  Aid.  The  situation  can  often  be  studied  to  ad- 
vantage from  a  drawing.  See  §  12. 

ILL.  3.  T.  "An  article  sold  for  $8  at  a  gain  of  J,  what  was  the 
cost? 

"  'Sold  at  a  gain  of  |'  means  '  the  selling  price  is  the  cost  plus  \  the 
cost.'  Draw  a  diagram  to  represent  the  terms. 

....- SP 


—  COST '"•GAIN"" 

"  Solve.  If  the  gain  is  \  the  cost,  what  is  the  selling  price?  f  the 
cost.  If  4  the  cost  is  $8,  what  is  \  the  cost?  $2.  If  \  the  cost  is  $  2, 
what  is  |  the  cost?  $6." 

ILL.  4-  T.  "  A  is  6|  %  taller  than  B.  B  is  how  many  per  cent 
shorter  than  A  ?  " 

"  What  does  this  mean  ?  A's  height  is  B's  height  plus  ^  B's  height. 
B's  shortage  is  what  part  of  A's  height?  Draw  a  diagram  to  repre- 
sent the  terms. 


"  Solve.  If  B's  shortage  is  ^  B's  height  and  A's  height  is  ||  B's 
height,  what  part  of  A's  height  is  B's  shortage  ?  -fa  or  6^  %." 

When  drawings  are  used,  fractions  may  usually  be 
omitted  in  the  solutions. 

Thus  :  In  ILL.  3,  If  4  parts  are  $  8,  how  much  is  1  part  ?  $2.  If  1 
part  is  $  2,  how  much  are  3  parts  ?  <f  6. 

38.  Into  Complex  Problems.  While  a  complex  problem 
can  always  be  separated  into  simple  problems,  it  may 
sometimes  be  separated  to  advantage  into  complex  prob- 
lems. 

ILL.     At  what  rate  will  $  200  gain  f  24  in  2  yr.  at  simple  interest  ? 
What  is  the  interest  of  $  200  for  2  yr.  at  1  %  ?    f  4.     If  the  interest 
is  $  4  at  1  %,  at  how  many  per  cent  is  the  interest  $  24  ?     6. 


§39    LESSON  8.    COMPLEX  PROBLEMS -BY  ARITHMETIC     33 

39.  Abbreviated  Solutions.  After  pupils  understand 
that  complex  problems  must  be  separated  into  component 
problems,  they  may  cast  each  component  problem  and  its 
answer  into  a  statement  more  or  less  abbreviated  instead 
of  into  question  and  answer. 

ILL.  If  the  cost  of  1  Ib.  is  12  ^,  and  the  selling  price  is  25  ?,  the  gain 
on  1  Ib.  is  13  f.  If  the  gain  on  1  Ib.  is  13  f>,  the  gain  on  10  Ib.  is  $  1.30. 
Or,  still  shorter,  the  gain  on  1  Ib.  is  13  p,  the  gain  on  10  Ib.  is  $  1.30. 

It  is  often  best  in  problems  involving  only  multiplica- 
tion and  division  to  express  the  answers  to  component 
problems  without  performing  the  operation. 

ILL.  What  is  the  simple  interest  of  $  720  for  157  days  at  7  %  ?  Use 
cancellation  method. 

P,  9  720  1.57 

T,  157  da.  2  14 

R'7%  X#x-?-x    1x157          628 

I,    ?  100    Jm  157 


I,    $21.98  cms.  21.98 

Multiplying  $  720  by  r ^  gives  the  interest  for  1  yr.  at  7  %  ;  dividing 
by  360,  for  1  da. ;  multiplying  by  157,  for  157  da. 

40.  Model  Analysis.      The  model  analysis  of  a  complex 
problem  consists  of  the  model  analyses  of  the  simple  prob- 
lems into  which  the  complex  problems  may  be  separated. 
Its  use  is  not  recommended  because  of  verbiage. 

ILL.  Since  the  cost  of  2  apples  is  6  p,  the  cost  of  1  apple  is  £  of  6  ^ 
or  3  f.  Since  the  cost  of  1  apple  is  3  ^,  the  cost  of  5  apples  is  5  times 
3  f  or  15  p. 

41.  Exercises.     1.  A,  B,  and  C  eat  8  loaves  of  bread,  each  the 
same  amount;  A  furnishes  3  loaves  and  B  5  loaves;  C  pays  24^  for 
what  he  eats.     How  much  should  A  receive  ?    Make  the  analytic  dia- 
gram.    2.  State  and  solve  the  simple  problems.    3.  Give  the  model 
analysis.     4.  A  buys  a  chair  and  a  table  for  $  35 ;  the  cost  of  the  chair 
is  f  of  the  cost  of  the  table.     What  is  the  cost  of  each  ?    Solve  by  the 
aid  of  a  drawing :  (a)  using  fractious ;  (6)  not  using  fractions. 


LESSON   9.     WRITTEN  PROBLEMS  —  ARRANGEMENT 


42.  Arrangement.      In   order   to   grasp   the   situations 
involved  in  a  problem,  it  is  helpful  to  write  what  is  given 
and  what  is  required.     Since  the  answer  to  each  com- 
ponent problem  is  to  be  used  in  the  solution  of  another,  it 
is  helpful  to  write  each  answer  as  soon  as  it  is  found. 
The   work   which    cannot    be    performed    mentally   may 
appear  at  the  right.     No  denominations  need  appear  in  the 
scratch  work  if  they  are  kept  in  the  statements. 

Write  as  concisely  as  possible  what  is  given  and  what  is 
required  in  a  vertical  line  with  a  short  horizontal  line 
below,  expressing  the  denominations.  Write  below  the 
line  the  answer  to  each  component  problem  as  soon  as  it  is 
found,  expressing  the  denominations.  At  the  right  put 
all  work  that  is  not  performed  mentally,  omitting  the 
denominations.  After  the  answer  write  ans. 

43.  Simple  Problems. 


Addition 

Subtraction 

ApT,   38 

Had,      $358 

Pr  T,   96 

Spent,    $299 

T,           ? 

Left,          ? 

T,        134  ans. 

Left,         $  59  ans. 

Multiplication 

Quotition 

C    1  H,  $216 

216 

C  1  H,  $216                 19 

C  19  H,       ? 

19 

En  C,  $4104      216)1I()4 

C  19  H,  $4104  ans. 

1944 

H,              ?                216 

216 

H,            19  ans.         1944 

4104 

1944 

34 


§44    LESSON  9.     WRITTEN  PROBLEMS  — ARRANGEMENT      35 


Partition 


C  19  H,  $4104 

C  1H, ? 

C1H, 


216 

19JH01 
$216  ans.      JJ8_ 
etc. 


NOTE.  Observe  the  awk- 
wardness of  retaining  the 
denominations  in  division. 

Quolition          Partition 
19  $516 

$216)$4104       19)$4104 


Explanations.  The  author  prefers  question  and  an- 
swer. See  §  34' 

Scratch  Work.  In  addition  and  subtraction  it  is  un- 
necessary to  rewrite  the  terms.  In  multiplication  and 
division  it  is  unnecessary  to  use  the  denominations  because 
they  appear  at  the  left.  To  require  the  denominations  is 
to  require  what  is  never  done  in  practice,  to  insist  upon 
distinctions  which  are  wearisome,  and  to  spend  energy 
which  may  be  better  applied.  See  the  note  above. 

44.  Complex  Problems.  1.  A  man  gave  $5760  cash 
and  128  cows  at  $56  in  exchange  for  land  at  $64  an  acre. 
How  many  acres  did  he  get  ? 

Pd,  $  5760  and  128  202 

56 


128  C  @  $56 
1  A,  $64 
A,        1_ 
Cows,  $  7168 
Land,  $12928 
Acres,     202  ans. 


768 

640 

7168 

5760 

12928 


64)12928 
128 
128 
128 


EXPL.  What  is  the  value  of  128  cows  @$56?  $7168.  If  the 
value  of  the  cows  is  $7168  and  the  cash  $5760,  what  is  the  .value  of 
the  land?  $12928.  If  the  value  of  1  A.  is  $64, how  many  acres  for 
$12928?  202. 


36      LESSON  9.    WRITTEN  PROBLEMS  — ARRANGEMENT     §45 

2.    What  is  the  reading  of  the  centigrade  thermometer 
when  the  reading  of  the  Fahrenheit  is  50  °  ? 

Fr.  C,  0°;  F,  32° 

BoilC,  100°;  F,  212r  F      c 

F  50°,   C  ?  "H  BOIL  (-'«> 


180°  F,  100°  C 

1°  1?      5°  P 
*>    9      ^ 

Ab.  Fr.,  18° 
18°  F,  10°  C  ana. 


so- 


X- 


FR 


EXPL.  If  the  freezing  point  of  F  is  32°  and  the  boiling  point  212°, 
•what  is  the  difference  ?  180°.  If  the  freezing  point  of  C  is  0°  and 
the  boiling  point  100°,  what  is  the  difference?  100°.  If  180°  F  is 
100°  C,  what  is  1°  F  ?  |°  C.  If  the  freezing  point  of  F  is  32°  and 
the  reading  is  50°,  how  much  above  freezing  is  the  reading?  18°,  etc. 

45.  Teaching.     In  writing  what  is  given  and  what  is 
required,  pupils  are  prone   to   spend   too   much   time  in 
expressing  terms  fully.      To  counteract  this  it  is  well  for 
the  teacher  to  read  a  problem  at  ordinary  speed,  while  no 
one  writes,  that  the  pupils  may  understand  it  as  a  whole ; 
and  then  to  read  it  slowly,  requiring  every  one  to  finish 
writing  as  soon  as  he  finishes  reading.     The  object  is  to 
be  as  concise  as  possible.     Another  good  plan  is  for  the 
teacher  to  read  the  problem  just  as  he  wishes  the  pupil  to 
write  it,  and  then  to  call  upon  some  one  to  state  it  in 

full. 

) 

46.  Proofs.     Pupils  should  not  be  allowed  to  consider 
a  solution  complete  until    they  have  proved  the  answer 
both  approximately  and  exactly. 

Approximate.  Consider  whether  the  answer  is  rea- 
sonable. 

ILL.  Problem,  §  44.  200  A.  at  $60  would  be  worth  $12000; 
202  A.  must  be  approximately  correct. 


§47     LESSON  9.    WRITTEN  PROBLEMS  — ARRANGEMENT      37 

Exact.  There  are  four  methods  :  First,  review  the 
work  with  care.  This  is  the  common  method;  it  has  been 
used  in  each  of  the  foregoing  problems.  Second,  solve 
the  problem  in  a  different  way.  This  is  of  value  when 
the  problem  can  be  separated  into  different  sets  of  com- 
ponent problems.  Third,  discover  whether  the  answer 
meets  the  conditions  of  the  problem.  This  is  of  value 
when  the  problem  is  algebraic  in  nature.  Fourth,  form 
and  solve  a  second  problem  in  which  the  required  term 
of  the  first  is  made  a  given  term  of  the  second,  and  some 
given  term  of  the  first  is  made  the  required  term  of  the 
second.  This  method  is  cumbrous  for  practice,  but  valua- 
ble as  an  exercise  for  the  pupil.  Observe  how  the  simple 
problems  of  §  43  in  multiplication,  quotition,  and  partition 
prove  each  other. 

ILL.  Fourth  Method.  T.  "From  the  solution  in  §44,  make 
another  problem  in  which  the  number  of  acres  shall  be  given  to  find 
the  cash  payment.  A  man  gave  128  cows  at  $  56,  and  a  sum  in  cash 
for  202  A.  of  land  at  $64  an  acre.  How  much  cash  did  he  pay?" 

47.  Discussion.      For  written  work,   the   requirement 
through  all  the  grades  of  writing  what  is  given  and  what 
is  required,  is  of  supreme  importance,  because  it  fixes  the 
attention  upon  the  situations.     The  time  lost  in  making 
the  statements  is  usually  more  than  gained  in  determining 
the  operations. 

48.  Exercises.      1.   At  396   Ib.    a   day,   a  ship's   crew  consume 
11088  Ib.  of   beef  in  28  da.     Solve  the  problem  arising  from  the 
omission  of  11088  Ib.     2.   From  the  omission  of  396  Ib.     3.    From 
the  omission  of  28  da.    4.   From  the  land  problem  in  §  44,  make  and 
solve  the  complex  problem  by  the  use  of  202  A.  as  a  known  term, 
and  the  number  of  cows  as  the  required  term.    5.  Solve  the  last 
problem  in  §  46.     6.   Show  that  your  answer  is  approximately  correct 
7.   Explain  your  proof  that  the  answer  is  exactly  correct. 


LESSON  10.     PROBLEMS  —  BY  ALGEBRA 


49.  Solutions  by  Algebra.  A  solution  by  algebra 
differs  from  a  solution  by  arithmetic  in  two  respects.  By 
algebra,  the  unknown  terms  are  represented  by  #,  y,  2 ; 
by  arithmetic,  they  are  expressed  in  full.  By  algebra,  the 
equations  are  solved  by  the  laws  of  algebra ;  by  arith- 
metic, they  are  solved  by  analysis. 

ILL.     An  article  is  sold  for  $  60  at  a  gain  of  £.     What  is  the  cost  ? 


Algebra 

S,  $  60         Let  x  =  the  cost 


c,  ? 


-  =  the  gain 


Qx 


C,  $  50  ans.     —  =  60 
5 

6*  =  300 
x  =  50 


Arithmetic 


S,  $60 
C,J?_ 

C,  $50  ans. 


The  difference  up  to  the  formation  of  the  equation  is  the  use  of  x 
in  the  one  and  of  cost  in  the  other.  By  algebra  the  equation  is  solved, 
Clearing  of  fractions,  6  x  =  300 ;  dividing  by  the  coefficient  of  x, 
x  =  50.  By  arithmetic  the  equation  is  solved,  If  f  of  the  cost  is  $  60, 
what  is  £  of  the  cost?  $10.  If  i  of  the  cost  is  $10,  what  is  f  of  the 
cost?  $50. 

Use  in  Arithmetic.  Solutions  by  algebra  are  valuable 
for  the  indirect  cases  in  percentage  and  interest,  and  for 
all  other  cases  in  which  an  operation  must  be  performed 
upon  the  required  term. 

38 


§50  LESSON   10.    PROBLEMS  — BY  FORMULA  39 

50.  Formulas.  The  relation  of  the  required  term  of 
a  problem  to  the  given  terms  may  be  expressed  by  an 
equation  in  which  the  required  term  is  the  left-hand 
member  and  combinations  of  the  given  terms  the  right- 
hand  member.  In  solutions  by  formula,  the  first  step  is 
to  get  the  formula  ;  it  must  be  recalled  from  memory,  it 
must  be  taken  from  a  book,  or  it  must  be  derived.  The 
second  step  is  to  substitute  the  given  values  and  to  solve 
the  equation. 

ILL.  1.  What  is  the  circumference  of  a  circle  whose  radius  is 
6  in.  ?  Recall  the  formula. 

R,  6  in.                                                                  3.1416 
C,?_  C  =  2  TT  fl  12 

C,  37.6992  in.  ant.  37.6992 

ILL.  2.  How  far  will  a  body  fall  from  rest  in  2  sec.?  Get  the 
formula  from  physics. 

T,  2  sec. 

D,  ?_  D  =  16-^  x  T» 
D,  64J  ft.  ans. 

ILL.  3.     Problem  of  p.  7,  case  6. 

A,  53  sh.                        (1)  P  =  B  x  R  R  =  A  -  D 

D,  47  sh.                        (2)  A  =  B  +  P  ~  A  +D 

R,?_                            (3)Z>^B-P  ^53-47 

R,  6%.  ans.                        2B  =  A  +  D  53+47 

2P  =  A-  D  =6% 

PROOF.  Second  Method  (§  46).  If  the  original  no.  plus  the 
no.  purchased  is  53,  and  the  original  no.  minus  the  no.  purchased 
is  47,  what  is  twice  the  original  no.  ?  100.  What  is  twice  the  no. 
purchased  ?  6.  If  the  original  no.  is  50,  and  the  no.  purchased  3, 
what  is  the  rate?  6%. 

Use  in  Arithmetic.  Formulas  may  be  used  in  problems 
in  which  the  relations  of  the  terms  must  be  found  by 
geometry,  by  physics,  or  by  involved  processes. 


40  LESSON   10.    PROBLEMS  —  BY  RULE  §51 

51.  Rules.  A  rule  is  the  translation  of  a  formula  into 
language  free  from  algebraic  expressions. 

ILL.  1.  C  —  2  TT  r  is  translated,  Given  the  radius  of  a  circle  to 
find  its  circumference,  multiply  twice  the  radius  by  3.1416. 

ILL.  2.  D  =  16  fa  x  T2  is  translated,  Given  the  time  of  a  body 
fallen  from  rest  to  find  the  distance,  multiply  16^  ft.  by  the  square 
of  the  number  of  seconds. 

A  —D 
ILL.  3.     R  =  —  —  -  is  translated,  Given  the  amount  and  the  differ- 


ence  to  find  the  rate,  divide  the  difference  between  the  amount  and 
difference  by  the  sum  of  the  amount  and  difference. 

The  teacher  will  find  the  exercise  of  translating  from 
formula  to  rule  excellent  to  strengthen  his  command  of 
expression. 


ILL.     A  =  V  s(s  —  a)(s  —  b)(s  —  c).     Translate. 

Given  the  sides  of  a  triangle  to  get  the  area,  find  the  continued 
product  of  the  half  sum  of  the  three  sides  and  the  remainders  found 
by  subtracting  each  side  from  the  half  sum  separately,  and  extract 
the  square  root  of  the  result. 

Use  in  Arithmetic.  Formerly,  rules  were  used  in  solv- 
ing most  of  the  problems  of  arithmetic.  At  present,  their 
use  is  restricted  for  the  most  part  to  mensuration.  The 
rule  is  stated  from  memory  and  its  directions  are  followed. 

ILL.     What  is  the  radius  of  a  circle  whose  area  is  78.54  sq.  in.? 

A.,  78.54  sq.  in.  25. 

R,      _  ?  3X1416.)78X5400. 

R,  5  in.  ans.  62832  . 

157080 
157080 

Given  the  area  of  a  circle  to  find  the  radius,  divide  the  area  by 
3.1416  and  extract  the  square  root  of  the  quotient. 


§51  LESSON   10.    PROBLEMS  — BY  PROPORTION  41 

52.  Mult,  and  Div.  Probs.     Problems  which  involve  no 
other  operation  than  multiplication  and  division  are  made 
up  of  a  number  of  different  terms  and  have  two  values  for 
each  term. 

ILL.  If  3  men  can  pick  240  bbl.  of  apples  in  8  da.,  how  many 
men  will  be  required  to  pick  480  bbl.  in  4  da.?  The  terms  are  men, 
barrels,  days.  The  values  for  men  are  3  and  x ;  for  barrels,  240  and 
480 ;  for  days,  8  and  4. 

53.  Relation  of  Terms.     The  relation  of  two  terms  is 
ascertained  by  multiplying  one  of  them  by  a  number  and 
noting  the  effect  upon  the  other. 

Men  and  Work.  What  is  the  effect  on  work  of  multiplying  no. 
of  men  by  a  number?  To  multiply  the  work  by  that  number.  Thus, 
twice  the  no.  of  men  do  twice  the  work.  The  no.  of  men  is  propor- 
tional to  the  work,  or  the  no.  of  men  varies  as  the  work. 

Men  and  Time.  What  is  the  effect  on  time  of  multiplying  no.  of 
men  by  a  number?  To  divide  time  by  that  number.  Thus,  twice 
the  no.  of  men  require  half  the  time.  The  no.  of  men  is  inversely 
proportional  to  the  time,  or  the  no.  of  men  varies  inversely  as  the 
time. 

Area  and  Radius.  What  is  the  effect  upon  the  radius  of  multiply- 
ing the  area  by  a  number?  To  multiply  the  square  of  the  radius  by 
that  number.  Thus  A=TT  R*  and  2  A  =  IT  x  2  R2.  The  area  of  a 
circle  is  proportional  to  the  square  of  the  radius,  or  the  area  of  a  circle 
varies  as  the  square  of  the  radius. 

54.  Exercises.     1.  After  losing  a  third  of  his  sheep  a  man  had  166 
left.     How  many  did  he  have  at  first?         Solve  by  algebra.     2.  In 
what  time  will  a  sum  of  money  double  at  6  %  simple  interest?     Solve 
by  algebra.     3.     On  a  lever  the  weight  is  54  lb.,  the  power  is  12 'lb., 
and  the  power's  distance  from  the  fulcrum  is  9  in.     What  is  the 
weight's  distance?    Solve  by  formula.    4.  What  is  the  surface  of  a 
sphere  whose  radius  is  6  in.  ?     Solve  by  rule.    6.  What  is  the  relation 
of  the  distance  fallen  by  a  body  from  rest  to  the  time  in  seconds  ? 


LESSON  11.    PROBLEMS  —  BY  PROPORTION 

55.  Two-Term  Problems.  Find  the  relation  of  the 
terms  by  multiplying  one  of  them  by  a  number  and  not- 
ing the  effect  on  the  other,  and  form  the  proportion  indi- 
cated. 

ILL.   1.   At  3  for  5  <f>  how  many  apples  can  be  bought  for  30^? 

3  ap,  5  f>  The  no.  of  apples  is  proportional  to 

x  ap,  30?  their  cost. 

No.  ap,  18  ans.  3  :  x  =  5  :  30 

3  x  30      1ft 

x  = =  18 

5 

ILL.  2.  If  2  men  require  10  da.  for  a  job,  how  many  days  do  5 
men  require? 

2  men,  10  da.  The  no.  of  men  is  inversely  propor- 

5  menI_x  da.  tional  to  the  time. 

No.  days,  4  ans.  2 : 5  =  x :  10 

.-•xlfi.4 

ILL.  3.  If  the  area  of  a  circle  whose  radius  is  5  in.  is  78.54  sq.  in., 
what  is  the  area  of  a  circle  whose  radius  is  10  in.  ? 

5  in.,  78.54    sq.  in.  The  area  of  a  circle  is  proportional 

10  in.,  x        sq.  in.  to  the  square  of  the  radius. 

Area7314.16  sq.  in.  ans.  78.54  :x  =  52: 102 

78.54  x  100 


25 


=  314.16 


ILL.  4.   If  the  area  of  a  circle  whose  radius  is  5  in.  is  78.54  sq.  in., 
what  is  the  radius  of  a  circle  whose  area  is  314.16  sq.  in.? 

5  in.,    78.54  sq.  in.  The  area  of  a  circle  is  proportional 

x  in.,  314.16  sq.  in.  to  the  square  of  the  radius. 

Radius,  10  in.  ans.  78.54 : 314.16  =  52 :  x2 

314.16  x  25 


78.54       =10° 


42 


§56  LESSON  11.    PROBLEMS  — BY  PROPORTION  43 

56.  N-Term  Problems.      Problems  of   more  than   two 
terms  give  rise  to  compound  proportion.     This  subject  is 
usually  omitted  from  arithmetics. 

Find  the  relation  of  the  required  term  to  each  of  the 
other  terms  and  make  a  proportion  for  each  relation. 
ILL.     Problem  of  §  52. 

3  men,  240  bbl.,  8  da.  _  f  240  :  480 

x  men,  480  bbl.,  4  da.  3:*_j    4    .    g 

XT-  10  3  x  480  x  8      10 

No.  men  12    ans.  x  = =  12 

240  x  4 

The  no.  of  men  is  proportional  to  the  no.  of  barrels  and  inversely 
proportional  to  the  no.  of  days.  This  means  that  so  far  as  the  no.  of 
barrels  is  concerned  3  :  x  =  240 : 480 ;  so  far  as  the  no.  of  days  is  con- 
cerned, 3  :  x  =  4  : 8 ;  so  far  as  both  are  concerned,  3  :  x  —  240  x  4 : 480  x  8. 

57.  Proportional  Parts.     The  statement  that  a  whole  is 
divided  into  parts  proportional  to  given  numbers  means 
that  the  ratio  of  the  two  sums  equals  the  ratio  of  each 
part  of  the  first  sum  to  the  corresponding  part  of  the  sec- 
ond sum. 

ILL.   1.   Divide  1728  into  parts  proportional  to  3,  4,  5. 

12;  3,4,5  12:1728  =  3:z 

1728  ;  x,_y,  z  12  : 1728  =  4  : y  s^s-*^^*-^ 

Parts,  432,  576,  720  ans.     12:1728  =  5:z 

ILL.   2.   Divide  68  into  parts  proportional  to  |  and  |. 
H;A,A  17:68=  8:x  i 

68;    ar,_y  17:68  =  9:y  'Cialx1 — I 

Parts,  32,  36  ans. 

58.  Use  in  Arithmetic.     There  is  no  great  need  for  pro- 
portion to  solve  the  problems  of  arithmetic.      It  seems 
necessary,  however,  to  teach  the  subject  in  the  elementary 
schools  because  its  terms  are  so  often  used  in  common 
speech.     All  mechanical  methods  which  avoid  a  study  of 
relations  should  be  avoided. 


44  LESSON   11.    PROBLEMS  — BY  VARIATION  §59 

59.  Two-Term  Problems.     Find  what  one  value  of  a 
term  has  been  multiplied  by  to  give  the  other  value,  and 
multiply  or  divide  the  given  value  of  the  other  term  as  the 
relation  indicates. 

ILL.  1.  §  55.  5  ^  has  been  multiplied  by  6 ;  multiplying  cost  by  a 
number  multiplies  apples  by  that  number ;  3  apples  must  be  multi- 
plied by  6.  —  Ans.  18  apples. 

ILL.  2.  §55.  2  men  has  been  multiplied  by  f ;  multiplying  men 
by  a  number  divides  days  by  that  number ;  10  da.  must  be  divided  by 
f.  —  Ans.  4  da. 

ILL.  3.  §  55.  5  in.  has  been  multiplied  by  2  ;  multiplying  the  radius 
of  a  circle  by  a  number  multiplies  the  area  by  the  square  of  that  num- 
ber; 78.54  sq.  in.  must  be  multiplied  by  4.  —  Ans.  314.16  sq.  in. 

ILL.  4.  §55.  78.54  sq.  in.  has  been  multiplied  by  4;  multiplying 
the  area  of  a  circle  by  a  number  multiplies  the  radius  by  the  square 
root  of  that  number;  5  in.  must  be  multiplied  by  2.  —  Ans.  10  in. 

60.  N-Term  Problems.     Compare  each  of  the  terms  with 
the  required  term  as  in  two-term  problems. 

ILL.  §  56.  240  bbl.  has  been  multiplied  by  2 ;  multiplying  barrels 
by  a  number  multiplies  men  by  that  number ;  3  men  must  be  multi- 
plied by  2.  —  .4ns.  6  men.  8  da.  has  been  multiplied  by  \ ;  multiplying 
days  by  a  number  divides  no.  of  men  by  that  number ;  6  men  must 
be  divided  by  \.  —  Ans.  12  men. 

61.  Proportional  Parts.     Find  what  one  sum  has  been 
multiplied  by  to  make  the  other  sum. 

ILL.  1.  §  57.  12  has  been  multiplied  by  144 ;  multiplying  the  sum 
by  a  number  multiplies  each  of  the  parts  by  that  number ;  3,  4,  5, 
must  be  multiplied  by  144.  —  A  ns.  432,  576,  720. 

ILL.  2.  §  57.  j£  has  been  multiplied  by  48 ;  f  and  £  must  be  mul- 
tiplied by  48.  —  Ans.  32,  36. 

62.  Use  in  Arithmetic.     The  method  of  variation  is  of 
great  value  in  all  problems  which  involve  multiplication 
and  division  because  its  use  requires  strict  attention  to  the 


§63  LESSON   11.    PROBLEMS  — BY  VARIATION  45 

relations  of  the  terms.  It  is  of  special  value  for  problems 
in  mensuration  which  have  to  do  with  similarity.  Thus, 
A  is  6  ft.  tall;  it  is  proposed  to  make  his  statue  12  ft.  tall. 
A's  little  finger  is  3  in.  long;  to  paint  a  statue  of  A's  size 
costs  $2;  the  weight  of  a  statue  of  A's  size  is  1000  Ib. 
What  will  be  the  length  of  the  little  finger  of  the  statue  ? 
What  will  it  cost  to  paint  the  statue  ?  What  will  be  the 
weight  of  the  statue  ? 

A's  height  has  been  multiplied  by  2 ;  multiplying  a  linear  part  by 
a  number  multiplies  a  linear  part  by  that  number,  multiplies  a  surface 
part  by  the  square  of  that  number,  and  multiplies  a  solid  part  by  the 
cube  of  that  number.  2  in.,  the  length  of  the  little  finger,  must  be 
multiplied  by  2  ;  $2,  the  cost  of  painting  a  statue  of  A's  size,  must  be 
multiplied  by  the  square  of  2 ;  1000  Ib.,  the  weight  of  the  statue, 
must  be  multiplied  by  the  cube  of  2. 

63.  Problems  in  General.     All  problems  may  be  solved 
by  stating  and  solving  their  simple  problems,  by  algebra,  by 
formula,  and  by  rule.     Problems  involving  multiplication 
and  division  may  also  be  solved  by  proportion  and  by 
variation.     The  secret  of  success  by  each  method  is  to  grasp 
the  situations.     Arrangement  of  the  work  is  an  important 
factor. 

64.  Exercises.     1.  If  a  body  falls  from  rest  64|  ft.  in  2  sec.,  how 
far  will  it  fall  in  6  sec.?     The  distance  varies  as  the  square  of  the  time 
in  seconds.     Solve  by  proportion.     2.  Solve  by  variation.     3.  If  3  boys 
earn  $3  in  3  da.,  how  many  boys  will  earn  $  100  in  100  da.  ?     Solve  by 
stating  simple  problems.     4.  Solve  by  algebra.     6.  Solve  by  formula.    6. 
Solve  by  rule.     7.  Solve  by  proportion.     8.  Solve  by  variation.     9. 
State  with  reasons  which  method  you  prefer  for  No.  3. 


LESSON   12.     LESSON  PLANS 

65.  Preparation.  Before  conducting  a  class  exercise 
the  teacher  should  have  a  definite  plan.  He  should  know 
exactly  what  he  is  going  to  do  and  exactly  how  he  is  going 
to  do  it.  He  will  then  look  forward  with  pleasure  to  the 
exercise  and  will  know  at  its  close  what  changes  to  make 
the  next  time  he  presents  a  similar  exercise. 

In  making  the  plan  the  principal  things  to  consider  are: 

1.  The  object  and  scope  of  the  exercise. 

2.  The  logical  steps  demanded  by  the  subject. 

3.  The  knowledge  which  pupils  must  have  before  they  are 
ready  to  take  up  the  subject. 

4.  The  means  which  must  be  used  to  induce  the  pupils  to 
take  the  logical  steps. 

Object  and  Scope.  The  object  of  a  class  exercise  may  be 
to  develop  a  new  subject,  to  drill  upon  a  subject  previously 
developed,  or  to  determine  how  well  a  subject  has  been 
mastered.  This  gives  rise  to  development  exercises,  drill 
exercises,  and  test  exercises. 

By  scope  is  meant  where  a  subject  shall  begin,  where  it 
shall  end,  and  how  fully  it  shall  be  treated.  Important 
factors  are  the  degree  of  maturity  of  the  child,  the  time 
allowed  for  the  exercise,  and  the  requirements  of  the  course 
of  study. 

Logical  Steps.  By  logical  steps  are  meant  the  steps  that 
must  be  taken  in  the  order  of  their  dependence.  See  §10. 

Knowledge.  The  teacher  should  consider  not  only  what 
knowledge  pupils  must  have  before  they  are  ready  to  take 

46 


§66  LESSON  12.  LESSON  PLANS  47 

up  a  subject,  but  also  whether  they  actually  possess  this 
knowledge.  Otherwise  he  will  present  what  has  been 
presented  before  or  he  will  present  what  is  irrelevant. 

Means.  This  topic  demands  the  teacher's  principal 
study.  He  must  consider  the  teachings  of  psychology, 
logic,  and  experience  to  determine  how  the  mind  acts  ; 
the  teachings  of  school  management  to  fix  upon  class 
government  and  mechanical  movements  ;  works  on  meth- 
ods and  history  of  education  to  test  his  theories;  and  in 
fine,  everything  which  he  has  studied  bearing  upon  the 
subject  to  add  to  his  efficiency. 

66.  Development  Exercises.  These  exercises  have  to 
do  with  new  subject  matter. 

ILL.  Combinations  of  4's  in  Addition.     First  lesson. 

Object  and  Scope.  To  get  the  pupils  to  repeat  from  memory  the 
table  of  4's  iu  addition. 

Steps.  To  discover  the  results  of  the  new  combinations  by  count- 
ing objects,  to  repeat  the  table  with  the  objects  in  sight,  and  to  repeat 
the  table  with  the  objects  out  of  sight. 

Knowledge.     Counting  and  the  tables  of  1's,  2's,  and  3's. 

Means.  To  show  the  need,  state  that  there  is  frequent  occasion  to 
find  the  sum  of  4  and  each  of  the  digits,  and  that  it  saves  time  to 
memorize  the  results.  To  help  the  pupils  to  satisfy  this  need,  have 
them  write  the  combinations  which  they  already  know  by  numerals 
and  the  new  combinations  by  dots  so  arranged  that  their  number  may 
be  recognized  by  form,  have  them  find  the  sums  by  counting  the  dots, 
have  them  repeat  the  table  with  the  dots  in  sight,  and  have  them 
repeat  the  table  with  the  dots  out  of  sight. 


I    2   3 
444 


67.     Drill  Exercises.     Like  finger   exercises   upon  the 
piano,  drill  exercises  in  arithmetic  are  to  give  ability  to  do 


48  LESSON   12.    LESSON  PLANS  §68 

accurately  and  rapidly  what  can  already  be  done  less  accu- 
rately and  less  rapidly. 

ILL.     Combinations  of  4's  in  Addition.     Second  lesson. 

Object  and  Scope.  To  get  pupils  to  call  the  results  of  the  combina- 
tions of  4's  in  addition  arranged  in  miscellaneous  order,  with  accuracy 
and  at  the  rate  of  three  a  second. 

Steps.  To  call  the  results  from  memory,  and  in  case  a  result  is 
missed,  to  repeat  the  entire  table. 

Knowledge.     Ability  to  repeat  the  table  of  4's  from  memory. 

Means.  To  get  all  the  combinations  before  the  pupils,  use  the 
circle  device  and  the  device  of  writing  the  addends  of  each  combina- 
tion in  a  vertical  line.  To  get  speed,  tap  on  the  board  slowly  and 
require  the  pupils  one  by  one  and  also  in  concert  to  call  the  results  in 
unison  with  the  sounds,  gradually  increasing  the  speed  until  the  rate 
of  three  combinations  a  second  is  attained.  If  this  speed  is  not 
reached  the  first  day,  repeat  the  exercise  at  intervals  for  months  or 
even  terms.  For  additional  practice,  state  simple  problems  requiring 
the  answers  instantly  without  any  form  of  explanation. 

B,    9    « 
0 

7  7159380426 

,   Q  ,  4,  4,  4,  4,  4,  4,  4,  4,  4,  4 

68.  Tests.  The  setting  of  proper  tests  requires  much 
skill.  Not  only  must  the  object  and  scope  be  determined 
with  great  care,  but  also  the  means  must  be  studied  with 
unusual  attention.  An  hour  spent  in  the  preparation  of 
a  test  will  often  save  several  hours  in  the  grading  of  papers 
and  will  insure  a  better  measurement  of  the  ability  of  the 
pupils. 

ILL.  Oral  Test.  Combinations  of  4's  in  Addition.  A  lesson  after 
the  subject  is  well  mastered. 

Object  and  Scope.  To  determine  whether  each  pupil  can  call  the 
results  of  the  combinations  of  4's  in  addition  accurately  and  without 
hesitation. 


§69 


LESSON  12.  LESSON  PLANS 


49 


Steps  and  Knowledge.     Same  as  in  the  drill  exercise. 

Means.  A  pupil  should  be  able  to  read  the  results  as  rapidly  when 
they  are  expressed  by  the  combinations  as  when  they  are  expressed 
by  the  common  method.  Write  a  half  dozen  of  the  results  by  the  com- 
mon method,  and  then  all  of  the  combinations.  Require  each  pupil 
to  read  all  the  results  at  the  same  rate  he  reads  those  expressed  by 
the  common  method. 

8064293157 
12,  10,  11,  13,  9,  6,  4,  4,  4,  4,  4,  4,  4,  4,  4,  4. 

ILL.      Written  Test.     The  forty-five  Combinations  in  Addition. 

Object  and  Scope.  To  determine  how  rapidly  pupils  can  write  the 
results  of  the  45  combinations  in  addition. 

Steps  and  Knowledge.     Same  as  in  the  drill  exercise. 

Means.  Give  each  pupil  a  paper  on  which  the  45  combinations 
have  been  written  twice  in  miscellaneous  order,  and  require  him  to 
write  as  many  of  the  answers  as  possible  in  one  minute. 

Graphs.  Below  is  a  graph  showing  the  records  of  ten 
pupils  in  the  written  test. 


IZ34S6789IO 


This  graph  shows  that  pupil  No.  1 
wrote  60  correct  answers  in  a  minute ; 
pupil  No.  2,  65 ;  and  so  on. 


69.  Exercises.  1.  Plan  a  development  exercise  on  the  combina- 
tions of  4's  in  multiplication.*  2.  A  drill  exercise.  3.  An  oral 
test.  4.  A  written  test  on  the  45  combinations  in  multiplication. 
5.  Prepare  a  graph  showing  the  records  of  ten  pupils  in  the  multi- 
plication test.  No.  1,  60  answers  in  a  minute;  2,  80;  3,75;  4,  60; 
5,  72;  6,  85;  7,  80;  8,  90;  9,  84;  10,  50. 

44        44 

*  Suggestion.     1  4,  2  4-»,  3  4-B,  <  4  >  4*4 ,  .  .  . 


LESSON  13.     IN  GENERAL 

70.  Mental  and  Written.     There  is  a  tendency  on  the 
part  of  both  teacher  and  pupil  to  use  the  pencil  too  freely. 
The  greater  part  of  the  computations  made  by  persons  in 
business  and  persons  in  the  trades  and  professions  are 
mental.     Often,  the  use  of  a  pencil  is  a  sign  of  weakness. 
The  first  step  in  the  solution  of  problems  is  to  grasp  the 
situation ;   the  second,  is  to  perform  the  operations.     In 
the  schoolroom  at  least  80  %  of  these  computations  should 
be  mental. 

ILL.  Pupils  should  be  able  to  work  such  examples  as  the  follow- 
ing mentally : 

$7.38  +  $  9.64;  $10.50  -  $6.84;  $7.85x12;  $57.65-12;  6f  +  8f; 
9f  -  4|;  49  x  f ;  49  -4-  |;  f  x  f ;  6f  x  8;  \  to  \  of  numbers  to  100; 
|  -  |;  3£  +  4J;  6%  of  500;  |%  of  $1000;  528  x  25;  1728  -  25;  ... 

71.  Economy  of  Time.     Operations  Easy.     Usually,  drill 
in  the  solution  of  problems  should  be  separate  from  drill 
in  the  performance  of  the  fundamental  operations.     In 
the  former,  the  numbers  should  be  such  as  not  to  distract 
the  attention  from  the  consideration  of  the  relations.     The 
greatest  care  must  be  exercised  that  problem  work  shall 
not  degenerate  into   an   exercise   in   multiplication   and 
division. 

ILL.  If  an  article  is  sold  for  $525.65  at  a  gain  of  27f  %  what  is 
the  cost? 

The  division  of  525.65  by  1.27^  is  so  difficult  that  the  attention  is 
likely  to  be  focused  upon  the  process  of  division  rather  than  upon  the 
discovery  that  division  is  necessary — $540  and  8%  are  better  numbers 
in  general. 

60 


§72  LESSON   13.    IN  GENERAL  51 

Operations  Omitted.  For  a  quick  review  of  complex 
problems,  pupils  may  be  asked  to  state  component  prob- 
lems and  to  name  the  answers  by  letters  without  com- 
putations. 

ILL.   In  what  time  will  $530.50  amount  to  $641.30  at  5%? 

If  the  principal  is  $  530.50  and  the  amount  is  $  641.30,  what  is  the 
interest?  fa.  What  is  the  interest  of  $530.50  for  1  yr.  at  5%? 
$  b.  If  the  entire  interest  is  $  a  and  the  interest  for  1  yr.  is  $  b  what 

is  the  number  of  years  ?    -. 
b 

Component  Problems  by  Diagrams.  For  the  discovery 
of  whether  pupils  understand  the  relations,  they  may  be 
required  to  prepare  analytical  diagrams  as  in  the  analysis 
of  sentences.  See  §  37. 

ILL.   Problem  above. 

. 
En  Interest 

Iforlyr. 

» 

72.  Expression.  By  the  Teacher.  Before  a  thought  is 
expressed  it  exists  in  the  mind  without  words  as  an  im- 
pulse. When  words  are  selected,  the  result  may  be  satis- 
fying to  the  speaker  but  unintelligible  to  the  hearer. 
Hence,  it  is  of  prime  importance  that  the  teacher  should 
use  with  accuracy  the  technical  terms  and  forms  of  phras- 
ing peculiar  to  each  subject.  Following  are  expressions 
having  a  tendency  to  arrest  development  which  the  author 
has  heard  from  teachers  in  the  schoolroom. 

It  is  criminal  for  a  teacher  to  give  long  development 
exercises  which  are  both  inaccurate  and  silly.  Hours  and 
hours  are  wasted  by  many  teachers  in  this  way,  and  pupils 
of  intelligence  come  to  despise  both  the  subject  and  the 


52  LESSON   13.     IN   GENERAL  §73 

teacher.     No  explanation  at  all  is  better  than  a  foolish  ex- 
planation. 

ILL.  1.  "  Tf  1  apple  costs  3  ^  the  cost  of  5  apples  will  be  as  many 
cents  as  5  multiplied  by  3^  or  15^."  Such  an  expression  has  a  ten- 
dency to  make  a  fool  of  the  pupil.  It  is  impossible  to  multiply  5  by 
3  ^ ;  the  phrasing  is  bad. 

ILL.  2  "  Since  12  is  f,  £  is  \  of  12."  12  =  f  is  a  false  statement, 
|  =  |  of  13  is  false. 

ILL.  3.  "  If  the  interest  of  1  yr.  amounts  to  f  12,  the  time  it  gains 
$  24  is  y^  of  24  which  is  2  yr."  '  The  interest  of  1  yr.'  is  ridiculous, 
'amount'  in  interest  problems  means  technically  'the  principal  plus 
the  interest,'  '  ^  of  24 '  is  2  and  not  2  yr.,  it  is  incorrect  to  take  -fa  of 
$  24,  for  then  the  denomination  of  the  answer  would  be  dollars. 

ILL.  4.  "  To  find  the  least  common  denominator  of  two  fractions 
by  inspection,  compare  the  successive  factors  of  the  largest  number 
with  the  smaller  until  a  factor  of  the  smaller  is  found."  The  correct 
expression  is,  "  Compare  the  successive  multiples  of  the  larger  denomi- 
nator with  the  smaller  until  a  multiple  of  the  smaller  is  found." 

By  the  pupil.  When  a  pupil  is  taking  up  a  new  topic, 
his  major  effort  is  to  master  the  thought.  He  should  not 
be  corrected  for  the  use  of  fragmentary  and  crude  ex- 
pressions until  quite  late.  For  a  time  it  is  enough  if  he 
hears  invariably  the  correct  forms  from  the  teacher.  He 
will  adopt  them  as  soon  as  he  gets  a  stronger  grip  upon 
the  thought.  He  should  not  be  required  to  give  the 
theoretical  explanations  of  the  fundamental  operations. 
He  will  understand  such  explanations  more  or  less  clearly 
if  they  are  skilfully  given  by  the  teacher,  but  he  has  not 
the  power  of  expression  to  make  them  himself. 

73.  Arrangement.  The  arrangement  of  the  work  is  of 
prime  importance. 


§74  LESSON   13.    IN  GENERAL  53 

The  teacher  should  always  paragraph  properly.  The 
rule  is  simple.  The  first  word  of  a  paragraph,  even  when 
it  is  a  numeral  or  a  letter,  should  begin  three  or  four 
letters  farther  to  the  right  than  the  first  word  of  each 
of  the  other  lines.  Thus  : 

[1.   Draw  two  vertical  lines.     Begin  the  first  word 

of    each    paragraph    on     the     right-hand    vertical.  • 

;2.   Begin  every  other  line  on  the  left-hand  vertical. 

If  a  drawing  is  used,  it  should  be  put  in  the  center  of 
the  page  from  right  to  left  and  no  writing  should  appear 
on  either  side. 

74.  Crutches.  The  teacher  makes  a  serious  mistake  if 
he  instructs  or  permits  pupils,  when  they  are  beginning  a 
topic,  to  write  figures  showing  how  many  are  to  be  car- 
ried, or  to  use  signs  of  operations  when  terms  are  written 
in  a  vertical  line,  because  in  many  cases  pupils  never  dis- 
card such  crutches.  Thus : 


68 
142 

The  small  figures  should  never  be  written,  even  by  the 
teacher  for  purposes  of  explanation.  Many  high  school 
graduates  who  enter  training  school  use  these  small  fig- 
ures habitually  in  performing  all  operations.  Nothing 
that  can  be  said  or  done  seems  to  be  effective  to  prevent 
them  from  perpetuating,  this  practice  when  they  become 
teachers. 

The  illustration  at  the  right  shows  how  one  of  these 
graduates  divides  by  7.  The  work  is  eloquent  of  im- 


7 

83 
-27 
56 

68 
x     9 
61'2 

7)378 
54 

7)3,22(46 
28 
42 
42 

54 


LESSON   13.    IN  GENERAL 


§75 


proper  instruction.  Not  only  does  she  use  crutches,  but 
she  also  employs  long  division  in  dividing  by  a  number 
of  one  order. 

The  use  of  the  signs  is  entirely  unnecessary,  is  opposed 
to  practice,  and  is  confusing  in  algebra. 

75.  Records.  Each  pupil  should  have  one  grade  and 
only  one  for  each  unit's  work  in  a  term,  and  the  name  of 
the  unit  should  be  written  in  connection  with  the  grade. 
The  average  of  a  great  number  of  grades  with  no  state- 
ment for  what  each  grade  was  given  is  of  little  value. 
One  grade  for  each  unit  together  with  the  name  of  the 
unit  gives  more  information  of  what  a  pupil  has  done 
during  a  term  than  a  hundred  miscellaneous  unmarked 
grades. 

ILL.  In  the  4  A  grade  (N.  Y.  City,  1912),  the  units  are  notation, 
counting,  addition,  subtraction,  multiplication,  division,  measure- 
ments, fractions,  problems.  Below  is  a  good  record  of  what  two 
pupils  have  accomplished. 


NAMES 

NOT. 

COUN. 

ADD. 

SUB. 

MULT. 

Div. 

MEAS. 

FK. 

PROB. 

Jovo^uyn/f  /r£/n/\At  Jo, 

a. 

a 

8 

€ 

8 

€ 

<g 

t 

'8 

(Ha/Ynfof  Tl'iaAsU  <3. 

a 

a 

a 

6 

a 

a 

8 

a 

a 

76.  Exercises.  1.  Give  the  correct  expression  for  111.  1,  §  72. 
2.  For  111.  2,  §  72.  3.  For  111.  3,  §  72.  4.  For  111.  4,  §  72.  6.  Criti- 
cise this  exercise  in  subtraction  of  fractions  and  rewrite  correctly : 


TEACHING  ARITHMETIC 


PART   II.     SUBJECT   MATTER 

LESSON  14.     NOTATION  AND  NUMERATION 

77.  How  Many.  Through  the  Ear.  The  first  need  in 
mathematics  is  to  express  how  many  there  are  in  a  group 
or  to  measure  a  group  with  reference  to  how  many.  The 
steps  are  the  same  as  in  all  measurement.  See  §  9. 

The  first  step  is  to  assert  that  there  are  as  many  in- 
dividuals as  in  a  well-known  group  and  to  name  the 
concept.  Thus,  there  are  as  many  as  there  are  fingers  on 
both  hands  or  ten. 

The  second  step  is  to  use  number  with  the  standard. 
Thus,  there  are  two  tens. 

The  third  step  is  to  select  larger  standards.  Thus, 
ten  tens  are  a  hundred  ;  ten  hundreds,  a  thousand  ;  a 
thousand  thousands,  a  million;  a  thousand  millions,  a 
billion  ;  and  so  on. 

The  fourth  step  is  to  use  two  or  more  standards.  Thus, 
there  are  *  two  tens  three,'  '  five  hundreds  seven  tens  six,' 
'three  hundred  twenty -five  millions  sixty  thousands  five.' 

66 


56  LESSON   14.    NOTATION  AND  NUMERATION  §78 

Through  the  Eye.  The  primary  method  is  to  display  as 
many  objects  as  there  are  individuals.  Thus,  tHI  tHJ  II. 

The  Arabic  method  of  writing  numbers  less  than  ten  is  to 
use  a  distinct  symbol  for  each  number.  Thus,  0,  1,  2,  •». 

The  Arabic  method  of  writing  numbers  less  than  a 
thousand  is  to  write  the  number  of  hundreds,  of  tens,  and 
of  units  by  the  plan  of  writing  numbers  less  than  ten  and 
to  express  the  names  of  the  groups  by  position.  Thus, 
536  means  '  5  hundreds  3  tens  6.' 

The  French  method  of  writing  numbers  greater  than 
a  thousand  is  to  write  the  number  of  thousands,  of  mil- 
lions, of  billions,  of  trillions,  and  of  higher  groups  by  the 
plan  of  writing  numbers  less  than  a  thousand  and  to  ex- 
press the  names  of  the  groups  by  position.  Thus, 
57,876,205  means  '57  millions  876  thousands  205  units.' 

78.  Through  Ten.  A  concept  is  more  vivid  than  its 
name.  Thus,  '  as  many  as  a  man  has  fingers  on  the  hand  ' 
is  more  vivid  than  'five.'  The  teacher  begins  with  the 
concept  but  passes  quickly  to  the  names  and  uses  them  in 
counting  objects,  motions,  and  sounds. 

ORAL.  T.  "  How  many  sticks  (//)  ?  There  are  as  many  sticks  as 
a  man  has  eyes;  one  for  this  eye  and  one  for  this  eye,  or  two.  How 
many  sticks  (///)  ?  Two  and  one  or  three.  How  many  sticks  (////)  ? 
As  many  as  a  horse  has  legs,  or  four." 

WRITTEN.  T.  "  How  shall  we  write  two?  We  will  make  a  mark 
for  each  individual,  //.  Take  pencil  and  paper.  Write  three,  four, 
five.  For  five,  /////  would  do,  but  we  will  draw  the  last  mark  across 
the  other  four  so  that  we  can  recognize  five  without  counting,  tHJ. 
Write  six,  seven ;  fNJ  /,  tW  II. 

"  There  is  a  shorter  way  of  writing  numbers.  To-day  we  will 
learn  the  short  way  of  writing  ////.  Upon  your  desk  there  is  a  paper 
upon  which  4  is  written  (each  part  about  a  half  inch  long)  and  upon 
the  board  there  is  also  a  4  (each  part  about  6  in.  long).  Trace  4  in 


§79  LESSON   14.    NOTATION  AND  NUMERATION  57 

the  air  (with  his  side  to  the  class  he  traces  and  the  pupils  imitate  his 
movements).  With  the  blunt  end  of  your  pencil  trace  the  4  on  your 
paper.  Begin  here.  Below  this  4,  with  the  sharp  end  of  your  pencil, 
write  another  4." 

79.  Through   a  Thousand.     The  teacher  has   at   least 
2000  sticks  about  4"  x  ^"  x  %"  and  a  box  of  short  rub- 
ber bands.     The  pupils  "put  the  sticks  into  bundles  of  ten, 
the  tens  into  bundles  of  a  hundred,  and  count  the  bun- 
dles and  single  sticks  left  over.     In  reading  and  writing 
in  the  abstract  they  visualize  these  objects. 

TENS  UNITS.  T.  "  Here  are  more  than  ten  sticks  (he  holds  64 
loose  sticks).  How  shall  we  find  how  many  there  are?  Take  them 
(he  distributes  the  sticks  and  rubber  bands),  put  them  into  bundles 
of  ten  with  a  band  around  each,  and  bring  to  me  the  bundles  of  ten 
and  the  single  sticks  left  over.  Now,  who  can  tell  how  many  sticks 
there  are?  6  tens  and  4.  Instead  of  '6  tens'  we  say  ' 6ty.'  What 
shall  we  call  7  tens?  8  tens?  9  tens?  5  tens?  Say^/7/ty,  not  fivety. 
4  tens?  3  tens?  Say  thirty,  not  threety.  2  tens?  Say  twenty,  not 
twoty.  Count  these  sticks  by  tens.  Ten,  twenty,  thirty,  .  .  .  How 
many  sticks  did  we  have  ?  Sixty-four. 

"Count  (he  takes  up  a  bundle  of  ten  and  one  stick  at  a  time). 
Ten,  ten-one,  ten-two,  ten-three,  .  .  .  ten-nine.  Instead  of  ten-one 
say  eleven;  instead  of  ten-two,  twelve;  instead  of  ten-three,  thirteen. 
What  shall  we  call  ten-four?  ten-five?  Say  fifteen,  not  fiveteen." 

T.  "How  shall  we  write  sixty-four?  We  might  write  6ty4;  we 
will  omit  the  ty  and  write  64.  Mary  may  take  from  the  box  eighty- 
three  sticks  (she  counts  ten,  twenty,  .  .  .  ).  You  may  alt  write 
eighty-three.  83.  What  is  understood  after  8  ?  Read  97.  What  is 
understood  after  9  ?  Jane  may  take  sixty  sticks  from  the  box.  You 
may  all  write  sixty.  Yes,  60 ;  Qty  naught." 

HUNDREDS  TENS  UNITS.    T.  "  Here  are  more  than  a  hundred  sticks 
(he  holds  385  loose  sticks) .    How  shall  we  find  how  many  there  are  ?  " 
The  teacher  proceeds  in  the  same  way  as  with  the  tens  and  units. 

80.  Beyond  a  Thousand.     French  Plan.     The  teacher 
has  drawings,  as  below,  upon  the  board  and  a  bundle  of 


58 


LESSON   14.    NOTATION  AND  NUMERATION 


§80 


a  thousand  sticks  shaped  as  a  4-in.  cube  upon  his  table. 
He  asks  the  pupils  to  visualize  ten  thousand  as  a  row  of 
10  such  bundles ;  a  hundred  thousand,  as  a  layer  of  10 
such  rows ;  and  a  million,  as  a  block  of  10  such  layers. 
He  asks  them  to  visualize  billions  and  the  higher  groups 
in  a  similar  way. 


ffff '(((((( 

THROUGH  A  MILLION.  T.  "  Here  are  more  than  a  thousand 
sticks.  How  shall  we  find  the  number?  Suppose  we  put  them  into 
bundles  of  a  thousand,  and  find  2  such  bundles  with  364  left  over. 
How  shall  we  write  this  number  ?  '  2  thousand  364 '  would  do,  but 
let  us  omit  thousand  and  write  2,364,  expressing  the  name  of  the 
group  by  position.  Write  '2  thousand  6.'  2,6  will  not  do  because  no 
place  is  left  for  hundreds  and  tens.  There  are  2  thousands  no  hun- 
dreds no  tens  6  units.  We  write  2,006.  That  is,  the  number  less 
than  a  thousand  must  be  written  with  three  figures.  Write  8  with 
three  figures.  Yes,  008.  Write  zero  with  three  figures.  Yes,  000, 
'  no  hundreds  no  tens  no  units.'  Write  968  thousand  16.  Yes,  968,016. 

"  How  can  you  picture  to  yourself  ten  thousand  ?  As  a  row  of  10 
bundles  of  a  thousand.  Show  me  on  the  floor  in  the  corner  of  the 
room  how  long  this  row  would  be  (40  in.).  How  can  you  picture  a 
hundred  thousand  ?  As  a  layer  of  10  such  rows.  Show  me  how  wide 
the  layer  would  be  (40  in.).  How  can  you  picture  10  hundred-thou- 
sand, or  a  million  ?  As  a  block  of  10  such  layers.  Show  me  how 
high  the  block  would  be  (40  in.)." 

HIGHER  GROUPS.  T.  "Suppose  there  are  many  more  than  a  mil- 
lion sticks,  how  can  we  represent  the  number?  Put  the  sticks  into 
bundles  of  a  thousand,  the  thousands  into  bundles  of  1000  thousands 
or  bundles  of  a  million,  the  millions  into  bundles  of  1000  millions  or 
bundles  of  a  billion,  and  so  on,  making  large  bundles  of  trillions, 
quadrillions,  quintillions,  and  so  on.  Picture  for  me  10  millions;  10 
ten -millions ;  10  hundred-millions  or  a  billion.  10  millions  would  be 


§81  LESSON   14.    NOTATION  AND  NUMERATION  59 

a  row  of  10  of  the  large  blocks ;  the  row  would  be  33^  ft.  or  longer 
than  the  width  of  this  room.  100  millions  would  be  a  layer  of  10 
such  rows ;  the  layer  would  be  33£  ft.  or  wider  than  the  length  of  this 
room.  A  billion  would  be  a  block  of  10  such  layers ;  the  height 
would  be  33 £  ft.  or  more  than  twice  the  height  of  this  room." 

T.  "Read  37000068782.  Beginning  at  the  right,  point  off  the 
number  into  periods  of  three  figures  each  (37,000,068,782).  Begin- 
ning at  the  right  numerate  by  periods  to  get  the  name  of  the  highest 
group  (units,  thousands,  millions,  billions).  Read  the  no.  of  each 
group  and  call  the  name  of  the  group  but  do  not  read  the  no.  of  a 
group  when  it  is  zero  (37  billion  68  thousand  782)." 

T.  "Write  37  billion  68  thousand  782.  Write  the  no.  of  the 
highest  group  and  express  the  name  of  the  group  by  position  (37,). 
Write  the  no.  of  the  next  highest  group,  using  three  figures,  and  ex- 
press the  name  of  the  group  by  position  (37,000,).  And  so  proceed 
(37,000,068,782)." 

81.  Uniformity.     The  above  plan  may  be  modified  for 
each  grade  to  suit  the  course  of  study.     Uniformity  of 
treatment  throughout  would  be  a  great  help  to  pupils. 

82.  Exercises.     1.   Teach  counting,  reading,  and  writing  of  hun- 
dreds tens  units  as  in  §  79.     2.    Teach  from  ten-thousand  through  a 
hundred-thousand.     3.    Teach   from    a    million    through   a   billion. 
4.   An  order  is  the  place  for  writing  one  figure ;  a  period,  the  place 
for  writing   three  figures.      Numerate   37,000,068 :    (a)  by   orders ; 
(6)    by  periods.     5.    Name    the   periods  forwards   and  backwards 
through  quintillions.     6.   Why  is  it  necessary  to  be  able  to  name 
these  periods  both  ways  ? 


LESSON  15.     NOTATION  AND  NUMERATION 

83.  Roman  Plan.  Numbers  are  written  by  means  of 
capital  letters.  For  units,  for  tens,  and  for  hundreds  the 
additive  principle  is  used  three  times  with  the  same  letter, 
then  the  subtractive  principle  with  a  new  letter,  and  then 
the  new  letter,  as  suggested  by  the  fingers  on  one  hand. 
(The  fingers  increase  in  length  from  the  little  finger  to 
the  forefinger  and  then  decrease).  The  additive  principle 
is  used  three  times  with  the  last  letter,  the  subtractive 
principle  with  a  new  letter,  and  the  new  letter,  as  sug- 
gested by  the  fingers  on  the  other  hand. 

This  method  requires  two  letters  for  units,  I  and  V; 
two  letters  for  tens,  X  and  L ;  and  two  letters  for  hun- 
dreds, C  and  D.  One  thousand  is  M ;  a  number  of  thou- 
sands is  the  number  under  a  horizontal  line. 

Units  Tens  Hundreds  Thousands 

II  X  C  Morl_ 

2  II  XX  CC  MMorll^ 

3  III  XXX  CCC          MMMorllT 

4  IV  XL  CD  IV 
5V                 L                 D                    _V 

6  VI  LX  DC  VI 

7  VII  LXX  DCC  VII 

8  VIII         LXXX         DCCC  VIII 

9  IX  XC  CM  IX 


68  million  539  thousand  754,  LXVIII  DXXXIX   DCCL1V. 

Through  Ten.  Pupils  should  be  taught  to  read  and 
write  numbers  as  suggested  by  the  additive  and  sub- 
tractive  principles  and  then  as  in  common  practice.  They 

60 


§83  LESSON   15.    NOTATION  AND   NUMERATION  61 

should  feel  the  necessity  of  a  new  letter  as  soon  as  the 
forefinger  is  reached. 


T.  "  On  the  face  of  a  clock  you  have  seen  numbers  written  by  the 
Roman  plan.  To-day  we  are  going  to  study  the  plan  through  ten. 
Touching  the  little  finger  of  your  left  hand  and  then  each  of  the 
other  fingers  count,  one,  two,  three,  one  from  five.  five.  Touching  the 
fingers  of  your  right  hand,  beginning  with  the  little  finger  count,  five 
and  one,  five  and  two,  five  and  three,  one  from  ten,  ten. 

"  Take  pencil  and  paper.  For  one,  write  capital  I ;  write  tico ; 
write  three.  Before  we  can  write  one  from  Jive  we  must  have  a  new 
letter  for  five ;  it  is  capital  V.  Write  one  from  Jive.  I  from  V  would 
do  but  we  omit  from,  IV.  Write  five ;  five  and  one,  V  and  I  would 
do  but  we  omit  and,  VI;  write  five  and  two:  five  and  three.  Before 
we  can  write  one  from,  ten  we  must  have  a  new  letter  for  ten ;  it  is 
capital  X.  Write  one  from  ten;  write  ten.  Read  I,  II,  III,  IV,  V, 
VI,  VII,  VIII,  IX,  X,  as  above ;  one,  two,  three,  one  from  five  .  .  . ; 
read  calling  the  usual  names  ;  one,  two,  three,  four  .  .  .  ." 

Through  a  Hundred.  Pupils  should  be  taught  to  develop 
tens  in  the  same  way  as  units.  In  writing  numbers  of 
two  orders  they  should  think  of  the  tens  as  suggested  by 
the  additive  and  subtractive  principles  and  then  of  the 
units  in  the  same  way. 

T.  "Touching  your  fingers  count,  one  ten,  two  tens,  three  tens, 
one  ten  from  five  tens,  five  tens ;  five  tens  and  one  ten,  five  tens  and 
two  tens,  five  tens  and  three  tens,  one  ten  from  ten  tens,  ten  tens. 
(He  proceeds  as  with  the  units.) 

"  Read  XL,  XC,  L,  C,  XXX,  LX  by  the  names  as  above;  by  the 
common  names.  What  shall  we  think  in  writing  eleven?  1  ten  1. 
Twelve  ?  1  ten  2.  Forty-eight  ?  4  tens,  5  and  3.  Sixty -nine  ?  5  tens 
and  1  ten,  1  from  ten." 


62  LESSON   15.    NOTATION  AND  NUMERATION  §84 

Through  a  Thousand.  Pupils  should  be  taught  to  de- 
velop hundreds  in  the  same  way  as  units. 

T.  "  Touching  your  fingers  count,  one  hundred,  two  hundreds, 
three  hundreds,  one  hundred  from  five  hundreds,  five  hundreds ;  five 
hundreds  and  one  hundred,  five  hundreds  and  two  hundreds,  five 
hundreds  and  three  hundreds,  one  hundred  from  ten  hundreds,  ten 
hundreds.  (He  proceeds  as  with  the  units.) 

"  What  shall  we  think  in  writing  974  ?  1  hundred  from  10  hun- 
dreds, 5  tens  and  2  tens,  1  from  5." 

Beyond  a  Thousand.  Pupils  should  be  taught  that  a 
bar  over  an  expression  multiplies  the  value  of  the  expres- 
sion by  1000. 

T.  "  For  1  thousand,  write  M  or  I ;  for  2  thousand,  MM  or  II ;  for 
3  thousand,  MMM  or  III.  How  shall  we  write  4  thousand  ?  IV.  17 


thousand  ?  XVII.    8  thousand  thousand  thousand  or  8  billion  ?  Vlll." 

84.  English  Plan.     Numbers  through  999  million  999 
thousand  999  are  read  and  written  by  the  English  plan 
exactly  the  same  as  by  the  French  plan.     The  next  num- 
ber greater  than  this  is  read  1000  million  by  the  English 
plan  and  1  billion  by  the  French  plan.     This  is  due  to 
the  fact  that  the  English  count  a  million  million  as  a  bil- 
lion, a  million  billion  as  a  trillion,  and  so  on,  while  the 
French  count  a  thousand  million  as  a  billion,  a  thousand 
billion  as  a  trillion,  and  so  on. 

ILL.  376,823456,025479,000006  is  read  376  trillion  823456  billion, 
25479  million  6.  For  reading,  numbers  must  be  pointed  off  into 
periods  of  six  figures  each ;  for  writing,  the  number  of  each  group 
except  the  highest  must  be  expressed  by  six  figures. 

85.  History.     The  Arabic  notation,  so  called  because 
the  Arabs  introduced  it  into  Europe,  was  used  by  the  Hin- 
dus as  early  as  200  B.C.    At  first,  the  names  of  the  groups 
were  represented  by  their  initial  letters,  but  about  400  A.D 
the  names  of  the  groups  were  expressed  by  position.     At 


§86  LESSON  15.    NOTATION  AND  NUMERATION  63 

this  time  the  symbol,  0,  was  introduced.  Thus,  from 
200  B.C.  to  400  A.D.,  six  hundred  fifty-eight  was  written 
6  h  5  t  8  u  ;  six  hundred  eight  was  written  6  h  8  u. 
Since  400  A.D.,  6  h  5  t  8  u  has  been  written  658 ;  6  h  8  u, 
has  been  written  608. 

The  Arabic  notation  was  introduced  into  Europe  in  the 
twelfth  century,  but  did  not  come  into  general  use  until 
the  sixteenth  century.  Nothing  was  added  to  what  the 
Hindus  practised  until  the  fourteenth  century.  Up  to 
this  time  no  single  word  expressed  more  than  a  thousand, 
and  the  best  way  of  reading  2,724,345,968,273  was  2  thou 
thou  thou  thou  724  thou  thou  thou  345  thou  thou  968 
thou  273.  The  French  suggested  that  a  thpusand  thou- 
sands be  called  a  million,  that  a  thousand  millions  be  called 
a  billion,  and  so  on.  The  Italians  suggested  that  a  thou- 
sand thousands  be  called  a  million,  that  a  million  millions 
be  called  a  billion,  that  a  million  billions  be  called  a  tril- 
lion, and  so  on.  Thus,  the  French  read  the  number  above 
4  2  trillion  724  billion  345  million  968  thousand  273,'  and 
the  English,  Germans,  and  other  nations  of  northern 
Europe,  read  it  '2  billion  724345  million  968273.' 

The  Roman  notation  is  supposed  to  have  been  invented 
by  the  Etruscans.  It  was  used  by  Europeans  exclusively 
until  the  twelfth  century  and  quite  generally  until  the 
fourteenth.  It  is  now  used  for  ornamental  purposes  only. 

86-  Exercises.  1.  Teach  Roman  numerals  from  X  through  L. 
2.  From  L  through  C.  3.  From  C  through  M.  4.  Beyond  M. 
5.  Why  is  the  French  plan  for  writing  numbers  superior  to  the 
Roman  for  practical  purposes?  6.  State  the  rule  for  reading  num- 
bers by  the  English  plan.  7.  State  the  rule  for  writing  numbers  by 
the  English  plan.  8.  Read  30000000287695400027;  (a)  by  the 
French  plan ;  (6)  by  the  English  plan. 


LESSON  16.     ADDITION 

87.  Needs.     A  whole  may  be  separated  into  parts,  each 
of  which  is  made  up  of  like  individuals.     Thus,  50  = 
3^  -+-  2^.     Let  us  classify  the  needs  which  arise  from  this 
statement  by  the  omission  of  each  term  in  succession  (§  7). 
If  the  whole  is  wanting,  the  requirement  becomes  No.  1 ' 
and  gives  rise  to  addition ;  if  one  of  the  parts  is  wanting, 
the  requirement  becomes  No.  2  and  gives  rise  to  subtrac- 
tion. 

1.  What  =  3^  +  2^? 

2.  5/  =  what  +  2^?  or  What  =  5^  -  2^? 

Addition  is  the  process  of  finding  the  number  of  indi- 
viduals in  the  whole  from  the  numbers  of  individuals  in 
the  parts.  The  whole  is  the  sum  or  amount,  the  parts 
are  addends. 

Subtraction  is  the  process  of  finding  the  number  of  indi- 
viduals in  one  of  two  parts  from  the  number  of  individ- 
uals in  the  whole  and  the  number  of  individuals  in  the 
other  part.  The  whole  is  the  minuend,  the  given  part  is 
the  subtrahend,  the  required  part  is  the  remainder. 

88.  Means.     The  primary  means  in  addition  is  to  take 
as  many  objects  as  there  are  individuals  in  each  addend 
and  to  find  the  result  by  counting. 

The  improved  means  is  to  find  the  sum  of  the  numbers 
in  each  order  separately,  by  calling  from  memory  the  sum 
of  two  digits,  by  increasing  the  result  by  a  third  digit,  by 
increasing  the  last  result  by  a  fourth  digit,  and  so  on. 

64 


§89  LESSON   16.    ADDITION  65 

This  necessitates  memorizing  the  sums  of  the  digits  taken 
two  at  a  time,  and  learning  how  to  increase  a  number  by 
each  of  the  digits. 

89.  Memorizing  the  Combinations.  The  combinations 
of  the  digits  taken  two  at  a  time  are  45  in  number : 

123456789  234567 

1,  1,  1,  1,  1,  1,  1,  1,  1  ;  2,  2,  2,  2,  2,  2, 
89  '3  456789  456789 

2,  2  ;  3,  3,  3,  3,  3,  3,  3 ;  4,  4,  4,  4,  4,  4 ; 

56789     6789  789     89     9 

5,  5,  5,  5,  5 ;  6,  6,  6,   6 ;  7,  7,  7 ;  8,  8 ;  9. 

The  steps  in  mastering  the  combinations  are  to  find  the 
sums  by  counting  and  to  memorize  the  results.  To  assist 
in  finding  the  sums,  the  domino  chart  is  valuable.  It 
separates  the  known  combinations  from  the  unknown, 
groups  the  individuals  in  such  a  way  that  their  number 
can  be  found  without  counting,  and  keeps  all  of  the  com- 
binations in  the  field  at  the  same  time.  To  assist  in 
memorizing  the  results,  the  combinations  and  sums  must 
be  repeated  as  found  in  the  tables  and  must  be  called 
miscellaneously.  The  sums  must  become  as  intimately 
associated  with  the  addends  as  the  names  of  objects  with 
the  objects  themselves.  There  must  be  no  counting  upon 
the  fingers  and  no  form  of  computation. 

A  pupil  should  not  be  regarded  as  proficient  until  he 
can  write  the  sums  accurately  at  the  rate  of  60  a  minute 
and  speak  them  accurately  at  the  rate  of  3  a  second.  To 
gain  these  ends  may  require  drills  at  intervals  during  the 
entire  school  course. 

ILL.     Combinations  of  4's.     Development  lesson.     For  plan  see  §  66. 
T.  "  To-day  we  are  going  to  find  the  sums  when  each  digit  is 
increased  by  four. 


66 


LESSON   16.    ADDITION 


§89 


"  A  boy  had  7  apples  and  obtained  4  more.  How  many  did  he 
then  have  ?  Use  sticks  for  apples.  Eleven  is  right,  but  it  takes  too 
long  to  proceed  in  this  way.  Who  can  suggest  a  shorter  way  ?  We 
will  find  the  results  and  then  memorize  them. 

"  You  may  draw  on  paper  three  rows  of  rectangles,  three  in  a  row 
(he  draws  them  on  board).  Let  us  write  all  the  combinations  of  4's, 
one  in  each  rectangle.  We  know  1  and  4,  2  and  4,  and  3  and  4,  and 
we  will  write  them  in  this  way  (teacher  and  pupils  fill  out  the  first 
row).  We  will  represent  4  and  4,  5  and  4,  ...  9  and  4  by  dots 
(teacher  and  pupils  fill  out  the  other  two  rows). 


"By  counting  Mary  may  find  the  sum  of  4  and  4,  5  and  4,  and 
so  on. 

"  You  may  all  look  at  your  charts  and  Jane  may  give  the  table  in 
this  way :  1  and  4  are  5,  2  and  4  are  6,  and  so  on.  Watch  carefully 
and  see  whether  she  makes  a  mistake. 

"  You  may  put  your  charts  in  your  desks  and  I  will  cover  mine. 
Who  can  give  the  table?  Henry  may  try.  He  missed  8  and  4. 
Look  at  my  chart  and  count  (the  teacher  uncovers  his  chart).  How 
many  are  8  and  4  ?  Henry  may  repeat  the  table  again  (the  teacher 
covers  his  chart).  The  class  together  may  repeat  the  table." 

ILL.     Combinations  of  4's.     Drill  lesson.     For  plan  see  §  67. 

T.  "  You  are  able  to  repeat  the  table  of  4's.  To-day  we  are  going 
to  call  the  results  in  miscellaneous  order  as  rapidly  as  possible. 

"  As  I  point,  Nellie  may  say,  '  9  and  4  are  13,  5  and  4  are  9,'  and 
so  on.  You  missed  7  and  4.  Give  the  table  of  4's.  How  many  are 
7  and  4? 


§  90  LESSON  16.    ADDITION  67 

"  As  I  point,  Susie  may  say,  '  13,  9,'  and  so  on. 

"  I  am  going  to  tap  upon  the  board  with  my  pointer  and  at  the 
word  now  I  want  John  to  say,  '  13,  9,'  and  so  on,  giving  a  result  at 
each  sound." 

Q,    9    * 
0 
7          7159380426 
•   8  -  4,  4,  4,  4,  4,  4,  4,  4,  4,  4. 

"  James  may  call  the  sums  keeping  in  time  with  the  sounds  (teacher 
taps  upon  the  board  at  a  uniform  rate).  Class  all  together,  the  same. 

"  Give  me  the  answers  to  these  problems  without  explaining.  If  I 
have  9^  and  get  4^,  how  much  shall  I  then  have?  If  I  have  7  $  and 
get  4^?  4^  and  get  4^?" 

ILL.     Combinations  of  4' s.     Oral  test.     For  plan  see  §  68. 
T.  "  I  wish  to  discover  whether  you  know  the  4's  as  well  as  you 
ought  to  know  them. 

8064293157 
12,  10,  11,  13,  9,  6,  4,  4,  4,  4,  4,  4,  4,  4,  4,  4. 

•'John  may  read  the  first  six  numbers  as  rapidly  as  possible. 
Read  the  whole  line  at  the  same  rate  as  the  first  six :  12, 10, 11, 13,  9, 
6,  12,  4,  10,  .  .  ."  If  John  slackens  in  speed  at  any  combination  his 
work  is  unsatisfactory. 

90.  Increasing  by  a  Digit.  In  grades  above  the  second 
year  pupils  should  be  drilled  at  intervals  throughout  the 
course  in  counting  from  1,  2,  and  so  on,  by  9's,  by  8's,  by 
7's,  by  6's,  by  5's,  by  4's,  by  3's,  and  by  2's.  A  pupil  is 
unsatisfactory  until  he  can  count  by  each  digit  as  rapidly 
as  by  1. 

ILL.  T.  "  Mary,  count  to  10  by  1's  as  fast  as  you  can.  Now,  be- 
ginning with  1  count  by  7's  to  99  at  the  same  rate.  You  count  by  7's 
much  more  slowly  than  by  1's.  I  want  you  to  practice  counting  by 
7's  several  times  a  day  for  a  week.  Class,  together,  count  by  7's  be- 
ginning with  1 ;  keep  up  with  the  pointer."  (The  teacher  beats  with 
the  pointer  up  and  down  at  a  rate  suited  to  the  slowest  in  the  class 
until  99  is  reached,  then  more  rapidly,  then  still  more  rapidly.) 


68  LESSON   16.    ADDITION  §91 

91.  Column  Addition.  A  pupil  should  never  be  allowed 
to  call  the  answer  to  an  example  in  addition  until  he 
knows  that  his  result  is  correct.  Speed  is  of  no  value 
without  accuracy.  A  clerk  who  sends  out  bills  with  in- 
correct footings  cannot  retain  his  position. 

Common  Check.  A  good  check  is  to  add  each  column 
from  the  bottom  up  and  from  the  top  down,  and  not  to 
add  the  next  column  until  the  sums  agree. 

ILL.  T.  "  We  will  take  an  example  in  addition.  Nellie  may 
work  at  the  board  and  others  on  paper. 


87 

48     28V 
69    27V 
74 
278 


Nellie's  Work 
28VV       29     SO 

27V 


"  Add  units'  column  from  the  bottom  up  and  place  the  sum  at  the 
right  (13,  21,  28)  ;  add  units'  column  from  the  top  down ;  if  the  sum 
is  the  same  as  before,  check  it;  if  not  the  same,  place  the  second  sum 
to  the  right  of  the  first  (15,  24,  28,  check).  Nellie,  I  see  that  you 
get  29.  Add  again  (she  gets  30).  Add  aloud  (at  last  she  gets  28 
several  times  and  checks  the  result  each  time).  Write  8  in  units' 
column. 

"Add  tens'  column  in  the  same  way  (9,  15,  19,  27)." 

Proof  by  Nines.  Expert  accountants  prove  addition  by 
excess  of  9's  occasionally  in  cases  of  perplexity.  Pupils 
may  practice  this  proof  from  the  sixth  year  on  if  time 
permits.  They  should  prove  the  following  principles 
inductively  (§  16). 

To  find  the  excess  of  9's  of  a  number,  add  its  digits, 
add  the  digits  of  the  sum,  and  so  proceed  until  the  last 
sum  is  less  than  9.  Count  9  as  0. 

To  prove  addition,  add  the  excesses  of  9's  of  the  ad- 


§92 


LESSON    16.    ADDITION 


69 


dends ;  the  excess  of  9's  of  the  result  should  be  the  excess 
of  9's  of  the  original  sum. 

9687  28V  3 

4832  26V  8 

6975  28V  0 

8369  29v  8 

29863  IV 

EXPL.     14,  21,  3;  12,  15,  17,  8;  13,  18,  0;  11,  17,  8;  8,  16,  19,  1; 
10,  16,  19,  IV. 

Double  Check.      Pupils   should   be    given   considerable 
practice  in  adding  numbers  arranged  as  in  statistics. 


1A 

IB 

2  A 

2B 

3A 

8B 

TOTAL 

It. 

32 

34 

29 

35 

36 

35 

T. 

52 

49 

48 

49 

51 

49 

W. 

28 

29 

24 

25 

25 

27 

T. 

33 

31 

32 

33 

33 

31 

F. 

22 

19 

26 

24 

24 

26 

Total 

ILL.  "  Fill  in  the  blanks,  adding  both  horizontally  and  vertically. 
For  a  double  check  add  the  totals  both  ways." 

• 

92.  Mental  Work.  Pupils  should  gain  the  ability  to 
increase  numbers  by  numbers  of  two  or  three  orders  men- 
tally. The  accomplishment  is  of  value  in  all  walks  of 
life.  Drills  should  be  given  with  the  addends  in  sight 
and  with  the  addends  out  of  sight.  See  §  70. 

587 
426 


T.  "  Find  the  sum  but  do  not  write  the  answer.     The  best  way  is 
to  begin  at  the  left.     Observe  that  5  must  be  increased  by  1  because 


70  LESSON   16.    ADDITION  §93 

f  is  more  than  9;  say,  6.     Observe  that  ?  must  be  increased  by  1 
because  |  is  more  than  9 ;  say,  4;  say,  0.     Find  the  sura  in  the  2d." 

T.  "Increase  $2.54  by  69?.      Say,  $2.54,  $3.14   ($2.54  +  60?), 
$3.23.     Increase  $5.67  by  $2.85.  —  Ans.  $7.67,  $8.47,  $8.52." 

93.  Written  Problems.     See  §  43. 

94.  Exercises.     1.    Give  a  development  lesson  on  the  combinations 
of  6's.     2.    A  drill  lesson.    3.   An  oral  test.     4.   Take  the  oral  test 
you  have  just  set.     Do  you  know  the  combinations  of  6's  as  well  as 
you  should  known  them?     6.    Take  the  test  of  §  90.     Can  you  count 
by  7's  as  easily  as  by  1's?    6.   Copy  the  example  of  §  91  under  double 
check  and  fill  in  the  blanks.     Explain  the  double  check.     7.   Prove 
inductively  the  first  principle  of  §  91,  under  proof  by  9's.     8.   Prove 
inductively  the  second  principle.     9.    Find  the  sum  in  the  example 
below  using  the  common  check.     10.   Prove  by  9's.     11.    Prove  by 
finding  the  sums  of  the  addends  in  sets  of  six  and  by  adding  the  sums. 

$786.95 

89.67 

437.69 

87.58 

6.39 

896.79 

88.77 

76.98 

97.86 

738.59 

46.73 

538.97 

9.87 

8.69 

238.49 

569.78 

49.99 

8.78 


LESSON  17.     SUBTRACTION 

95.  Needs.     The  need  of  subtraction  has  been  developed 
in  §  87.     It  manifests  itself  in  two  forms,  to  find  what 
must  be  added  to,  one  number  to  make  another,  and  to 
find  the  result  when  one  number  is  diminished  by  another. 
Thus,  5^  =  what  +  3^?     What  =  5^-3^? 

96.  Means.     The  process  of  finding  what  must  be  added 
to  one  number  to  make  another  is  well  illustrated  by  the 
process  of  making  change,  in  which  it  is  customary  to 
add  to  the  purchase  price  by  cents  until  a  multiple  of  five 
is  reached  and  then  to  add  by  multiples  of  five.     In  col- 
umn subtraction,  enough  is  added  to  the  subtrahend  to 
make  units'  digit  the  same  as  units'  digit  of  the  minuend, 
then  enough  is  added  to  the  result  to  makes  tens'  figure 
the  same  as  tens'  figure  of  the  minuend,  and  so  on.     This 
necessitates  ability  to  call  one  of  the  addends  in  each  com- 
bination in  addition  from  the  sum  and  the  other  addend. 

The  process  of  taking  one  number  away  from  another 
requires  the  changing  of  the  numbers  so  as  to  make  each 
digit  of  the  subtrahend  equal  to  or  less  than  the  corre- 
sponding digit  of  the  minuend.  This  necessitates  the 
mastery  of  the  combinations  in  subtraction  whose  minu- 
ends are  the  sums  of  the  combinations  in  addition  and 
whose  subtrahends  are  the  addends  of  the  combinations 
in  addition.  They  are  90  in  number. 

Pupils  should  be  drilled  upon  the  combinations  from 
both  points  of  view,  but  should  be  taught  only  one  method 
of  column  subtraction. 

71 


72  LESSON   17.    SUBTRACTION  §97 

97.  The  Combinations.  The  combinations  in  subtrac- 
tion are  known  as  soon  as  the  combinations  in  addition 
are  known  but  drill  is  necessary  to  accustom  the  mind  to 
the  new  phase.  The  pupil  is  unsatisfactory  until  he  can 
call  three  results  a  second. 

ILL.     Combinations  of  4's  in  Subtraction.     Drill  exercise. 

T.  "  We  are  going  to  drill  on  the  combinations  of  4's  in  subtrac- 
tion. Draw  a  circle;  on  the  outside  write  the  sums  of  the  combina- 
tions of  4's  in  addition ;  at  the  center  write  4. 


11  7  5  13  6   9  10  12  4  8 
4,4,4,    4,4,4,    4,    4,4,4. 

"  Henry,  read  from  the  circle  in  this  way,  4  and  7  are  11,  4  and  5 
are  9,  and  so  on.  Mary,  read  in  this  way,  11  minus  4  are  7,  9  minus  4 
are  5,  and  so  on.  You  missed  12  minus  4.  4  and  how  many  are  12  ? 
John,  read  in  this  way,  7,  5,  and  so  on.  As  I  tap  on  the  board  Susan 
may  call  the  results,  7,  5,  and  so  on  in  perfect  time  (he  taps  at  the 
rate  of  two  a  second).  Nellie,  read  the  remainders  at  the  right,  7,  3, 
and  so  on  in  perfect  time  (he  taps  at  the  rate  of  two  a  second). 

"Give  me  the  answer  without  explanation.  A  girl  has  4^  and 
wants  to  spend  13^ ;  how  much  does  she  need  ?  A  boy  had  12  $  and 
spent  4j£ ;  how  much  did  he  have  left?" 

98.  Column  Subtraction.  Austrian  Method.  It  is  well 
to  introduce  the  work  by  exercises  in  making  change. 

T.  "  On  each  desk  there  is  an  envelope  containing  a  toy  nickel  and 
five  toy  cents.  I  am  going  to  buy  things  of  you  and  ask  you  to 
make  change.  I  buy  a  top  for  4  p  and  pay  with  a  dime.  Make  the 
change  (they  all  do  it).  John,  make  the  change.  4  (he  names  the 
purchase  price),  5  (he  puts  down  a  cent),  10  (he  puts  down  a 
nickel)." 

T.  "  On  each  desk  there  is  an  envelope  containing  toy  money.  I 
buy  a  cap  for  59^  and  pay  with  a  dollar  bill.  Make  the  change 


§98  LESSON   17.    SUBTRACTION  73 

(they  all  do  it).  Make  the  change,  John.  59  (he  names  the  pur- 
chase price),  60  (he  puts  down  1^),  85  (he  puts  down  a  quarter), 
95  (he  puts  down  a  dime),  $  1  (he  puts  down  a  nickel).  How  much 
change  ?  41  ?(1  +  25  +  10  +  5)." 

T.  "  To-day  we  are  going  to  find  what  must  be  added  to  one 
number  to  make  another  or  to  subtract. 

89  83  8786 

37  59  3879 

"  James  and  Henry  at  the  board.  What  must  be  added  to  37  to 
make  89?  How  shall  we  proceed?  To  37  add  enough  to  make  a 
number  whose  units'  digit  is  9  or  to  7  enough  to  make  9.  Yes,  add 
2  (the  sum  is  39).  To  39  add  enough  to  make  89  or  to  3  tens  enough 
to  make  8  tens.  Yes,  add  5  tens.  AVhat  is  the  answer?  52. 

"A  harder  example.  What  must  be  added  to  59  to  make  83? 
How  shall  we  proceed  ?  To  59  add  enough  to  make  a  number  whose 
units'  digit  is  3  or  to  9  enough  to  make  1  ten  3.  Yes,  add  4  (the 
sum  is  63).  To  63  add  enough  to  make  83  or  to  6  tens  enough  to 
make  8  tens.  Yes,  add  2  tens.  What  is  the  answer  ?  24. 

"  Again.  This  time  say,  9  and  4  are  13  (write  4),  6  and  2  are  8 
(write  2).  Without  rewriting  add  59  and  24  to  see  if  the  sum  is  83." 

T.  "  From  8786  subtract  3879.  Say  9  and  7  are  16  (write  7),  8 
and  0  are  8  (write  0),  8  and  9  are  17  (write  9),  4  and  4  are  8  (write 
4).  Prove  your  answer." 

First  Italian  Method.  The  first  Italian  method  depends 
upon  the  principle  that  taking  one  from  any  order  and 
adding  its  equivalent  to  the  next  lower  order  does  not 
change  the  value  of  the  number. 

8006  799     (16) 

3879  387        9 

ILL.  The  numbers  are  changed  in  the  imagination  as  at  the  right 
to  make  the  digit  in  each  order  of  the  subtrahend  equal  to  or  less 
than  the  number  in  the  corresponding  order  of  the  minuend.  The 
pupil  says,  "  9  from  16,  7  ;  7  from  9,  2  ;  8  from  9,  1 ;  3  from  7,  4." 

Second  Italian  Method.  The  second  Italian  method 
depends  upon  the  principle  that  adding  10  to  any  order 


74 


LESSON   17.    SUBTRACTION 


§99 


of   the  minuend  and  1  to  the  next   higher  order  of  the 
subtrahend  cannot  affect  the  remainder. 


8006 
3879 


8     (10)     (10)     (16) 
4989 


ILL.  The  numbers  are  changed  in  the  imagination  as  at  the  right 
to  make  the  digit  in  each  order  of  the  subtrahend  equal  to  or  less 
than  the  number  in  the  corresponding  order  of  the  minuend.  The 
pupil  says,  "  9  from  16,  7 ;  8  from  10,  2 ;  9  from  10,  1 ;  4  from  8,  4." 

99.  Ledger  Balances.  The  Austrian  method  is  of  value 
in  finding  ledger  balances. 


36 

89 

225 

95 

38 

387 

50 

28 

67 

79 

58 

Balance 

371 

98 

612 

50 

612 

50 

Brought  forward 

371 

98 

ILL.  The  sum  of  the  larger  side  is  612.50.  We  want  to  find  what 
must  be  added  to  the  smaller  side  to  make  this  number.  15,  23,  32 
(sum  of  units),  40  (enough  must  be  added  to  32  to  make  a  number 
whose  units'  digit  is  0  or  to  make  40.  Write  8)  ;  9  (the  4  of  40  plus 
5),  15,  18,  26,  35  (write  9)  ;  and  so  on. 

A  check  on  the  work  is  to  add  both  sides ;  the  sums  should  be  the 
same. 

100.  Mental  Subtraction.  Pupils  should  be  drilled  until 
they  can  call  quickly  the  difference  between  numbers 
of  two  or  three  orders  and  numbers  of  two  orders,  and 
until  they  can  subtract  readily  numbers  of  two  orders 
without  use  of  a  pencil.  See  §  70. 

ILL.  Minuend  100.  Call  what  must  be  added  to  tens'  digit  to 
make  9  and  what  must  be  added  to  units'  digit  to  make  10.  Thus, 
100  -  37,  6ty  3 ;  100  -  58,  4ty  2. 


§101  LESSON  17.    SUBTRACTION  75 

ILL.  Minuend  less  than  100.  Call  what  must  be  added  to  tens' 
digit  or  tens'  digit  plus  one  to  make  tens'  digit  of  the  minuend,  and 
what  must  be  added  to  units'  digit  to  make  the  least  number  ending 
in  units'  digit  of  the  minuend.  Thus,  83  —  42,  4ty  1 ;  83  —  19, 
6ty4. 

101.  Written  Problems.     See  §  43. 

102.  Discussion.      The  three  methods  of   column  sub- 
traction have  been  used  since  200  B.C.     Two  of  them 
were  brought  ir-to   prominence  by  the   Italians  and  the 
other  by  the  Austrians.      Each  method  has  enthusiastic 
supporters.     The  advocates  of  the  Austrian  method  claim 
that  it  embodies  the  primary  notion  of  subtraction  and 
that  it  does  away  with  borrowing. 

103.  Exercises.     1.   Explain  the  derivation  of  the  terms,  addend, 
subtrahend,  minuend.     2.   Show  that  there  are  90  combinations  in 
subtraction  derived  from  the  combinations  in  addition.     3.   Set  an 
oral  test  for  determining  whether  a  pupil  knows  the  combinations  of 
6's   as  well  as  he  should.     4.   Try  the  test  yourself.     What  is  the 
result  ?    5.   Describe  a  written  test  for  determining  the  proficiency  of 
a  pupil  in  writing  the  results  of  the  90  combinations.     See  §  68. 
6.   Say  the  words  which  a  pupil  should  use  in  subtracting  76083456 
from   87692063:    (a)  by  the   Austrian   method;    (6)   by  the   First 
Italian ;  (c)  by  the  Second  Italian.     7.   State   with  reasons  which 
method  you  prefer. 


LESSON   18.     MULTIPLICATION 

104.  Needs.  A  whole  may  be  separated  into  parts  each 
of  which  is  made  up  of  the  same  number  of  like  individ- 
uals. Thus,  6^=  2^  +  2^  +  2^,  or  6^  =  2jz?  3  times,  or 
6^  =  3  times  2^,  or  6^  =  3  x  2£. 

Let  us  classify  the  needs  which  arise  from  the  statement 
by  the  omission  of  each  term  in  succession  (§  7).  If  the 
whole  is  wanting,  the  requirement  becomes  No.  1  and 
gives  rise  to  multiplication  ;  if  the  number  of  equal  parts 
is  wanting,  the  requirement  becomes  No.  2  and  gives  rise 
to  that  case  in  division  which  is  known  as  quotition  ;  if 
one  of  the  equal  parts  is  wanting,  the  requirement  be- 
comes No.  3  and  gives  rise  to  that  case  in  division  which 
is  known  as  partition. 


1.  What  =  3  x 

2.  6^  =  what  x  2^?  or  What  =  6^- 

3.  6j*  =  3  x  what?  or  What  =  6j*  -5-  3?  or  What  =  }  of  6^? 

Multiplication  is  the  process  of  finding  the  number  of 
individuals  in  the  whole  from  the  number  of  individuals 
in  one  of  the  equal  parts  and  the  number  of  the  equal 
parts.  The  whole  is  the  product  ;  one  of  the  equal  parts, 
the  multiplicand;  the  number  of  equal  parts,  the  multi- 
plier. 

The  multiplier  must  always  be  abstract.  '  3  x  2  ^  '  is 
read  *  3  times  2^  '  ;  '  2  ^  x  3  '  is  read  '2^  multiplied  by  3.' 

Quotition  is  the  process  of  finding  the  number  of  equal 
parts  from  the  number  of  individuals  in  the  whole  and 
the  number  of  individuals  in  one  of  the  equal  parts. 

76 


§105  LESSON   18.    MULTIPLICATION  77 

Partition  is  the  process  of  finding  the  number  of  individ- 
uals in  one  of  the  equal  parts  from  the  number  of  individ- 
uals in  the  whole  and  the  number  of  equal  parts.  Division 
is  the  process  of  finding  one  of  two  numbers  from  their 
product  and  the  other  number.  It  embraces  quotition 
and  partition.  The  whole  or  product  is  the  dividend; 
the  given  number,  the  divisor;  the  required  number,  the 
quotient. 

105.  Means.    Multiplication.     The   primary   method  is 
counting.     A  shorter  method  is  addition.     The  improved 
method  is  to  call  from  memory  the  products  of  the  digit 
in  each  order  of   the  multiplicand   by  the  digit  in  each 
order  of  the  multiplier,  and  to  add. 

106.  The  Combinations.     The  improved  method  necessi- 
tates memorizing  the  products  of  the   nine    digits  taken 
two  at  a  time.     They  are  45  in  number.     See  §  89.     The 
steps  in  mastering  the  combinations  are  to  find  the  prod- 
ucts by  addition  and  to  memorize  them. 

ILL.  Combinations  of  4's  in  Multiplication.  Plan  of  development 
exercise.  See  §  66. 

Object  and  Scope.  To  get  the  pupils  to  repeat  from  memory  the 
table  of  4's  in  multiplication. 

Steps.  To  discover  the  results  of  the  new  combinations  by  addi- 
tion, to  repeat  the  table  with  the  objects  in  sight,  and  to  repeat  the 
table  with  the  objects  out  of  sight. 

Knowledge.     Addition  by  4's  and  the  tables  of  1's,  2's,  and  3's. 

Means.  To  show  the  need,  state  that  there  is  frequent  occasion  for 
finding  the  product  of  4  and  each  of  the  digits,  and  that  it  saves  time 
to  memorize  the  results.  To  help  the  pupils  to  satisfy  this  need,  have 
them  write  the  combinations  which  they  already  know  in  this  way, 
1 4,  2  4-s,  3  4-s,  and  the  new  combinations  by  4's  so  arranged  that  their 
number  may  be  recognized  by  form,  have  them  find  the  sums  by  add- 
ing the  4's,  have  them  repeat  the  table  with  the  4's  in  sight,  and  have 
them  repeat  the  table  with  the  4's  out  of  sight. 


78 


LESSON   18.    MULTIPLICATION 


§107 


ILL.  Combinations  of  4's  in  Multiplication.  Plan  of  drill  exercise. 
See  §  67. 

Object  and  Scope.  To  get  pupils  to  call  the  results  of  the  combina- 
tions of  4's  in  multiplication  arranged  in  miscellaneous  order,  with 
accuracy  and  at  the  rate  of  two  a  second. 

Steps.  To  call  the  results  from  memory,  and  in  case  a  result  is 
missed,  to  repeat  the  entire  table. 

Knowledge.     Ability  to  repeat  the  table  of  4's  from  memory. 

Means.  To  get  all  the  combinations  before  the  pupils,  use  the 
circle  device  and  the  device  of  writing  the  terms  of  each  combination 
in  a  vertical  line.  To  get  speed,  tap  on  the  board  slowly  and  require 
the  pupils  one  by  one  and  in  concert  to  call  the  results  in  unison  with 
the  sounds,  gradually  increasing  the  speed  until  the  rate  of  two  com- 
binations a  second  is  attained.  If  this  speed  is  not  reached  the  first 
day,  repeat  the  exercise  at  intervals  for  months  or  even  terms.  For 
additional  practice,  state  simple  problems,  requiring  the  answers 
instantly  without  any  form  of  explanation. 


8205946173 
4,4,4,4,4,4,4,4,4,4. 


107.    Carrying.     The  improved  method  necessitates  in- 
creasing the  product  of  two  digits  by  a  number  from  one 


§  108  LESSON  18.    MULTIPLICATION  79 

to  a  number  one  less  than  the  multiplier.  For  a  time 
daily  drills,  and  later  drills  at  intervals  throughout  the 
course,  are  of  value. 

ILL.  Carrying  with  4's-  First  form  of  drill.  T.  "  Henry,  increase 
the  products  from  the  circle  (§106)  by  1.  Say,  37,  21,  1,  ... 
Mary,  increase  the  products  by  2.  Say,  38,  22,  2,  ...  Joseph,  in- 
crease the  products  by  3.  Say,  39,  23,  3,  ..." 

ILL.  Carrying  with  4's-  Second  form  of  drill.  T.  "  John,  begin- 
ning at  the  right  find  mentally  the  product  of  each  digit  by  4,  add 
mentally  the  tens  of  the  preceding  product,  and  speak  the  answers. 
Do  not  write  anything.  Thus,  32,  15,  29,  2,  ..." 

40738                               12659 
4  4 

108.  Multiplier  One  Order.  The  pupil  should  strive  to 
have  the  product  of  each  combination  recalled  and  the 
addition  made  subconsciously  as  in  the  second  form  of 
drill  in  the  last  paragraph.  After  sufficient  practice  it 
will  seem  to  him  that  he  is  simply  calling  the  results  of 
another's  work. 

ILL.  Multiplier  2.  T.  "  We  are  going  to  multiply  large  numbers 
by  2.  John,  at  the  board;  others,  pencil  and  paper.  Multiply 
768  by  2. 


Teacher's  Work 

768 
2 


1536 


John's  Work 

768 
768 


1536 


"John's  work  is  right.  768  x  2  means  to  find  the  sum  of  two 
768's.  I  want  you  to  use  a  shorter  way.  Write  768  and  below  it  2 
showing  that  there  are  two  768's.  Instead  of  adding  8  and  8,  think 
2  8's  are  16  and  say,  16;  instead  of  adding  6  and  6  and  1,  think  2  6's 
are  12,  and  1  are  13,  and  say  13 ;  and  so  on." 

ILL.  Multiplier  3.  T.  "  We  are  going  to  multiply  large  numbers 
by  3.  Multiply  768  by  3.  Proceed  as  in  multiplying  768  by  2. 
Prove  your  answer  by  going  over  the  work  a  second  time.  Every  one 
is  right.  I  am  greatly  pleased.' 


80 


LESSON  18.     MULTIPLICATION 


109 


109.  Multiplier  10.  To  multiply  by  10,  move  each 
figure  one  place  to  the  left.  This  seems  to  be  a  better 
rule  than  'annex  a  cipher,'  because  a  cipher  is  not  an- 
nexed in  multiplying  by  a  number  of  two  orders  and  be- 
cause annexing  a  cipher  does  not  multiply  a  decimal  by 
10. 

ILL.  T.  "  We  know  how  to  multiply  by  9.  How  shall  we  multiply 
by  10  ?  Let  us  multiply  23  by  10. 


HUNDREDS 

TENS 

UNITS 

2 

3 

2 

3 

"What  is  3  units  x  10?  3  tens.  What  is  2  tens  x  10?  2  hun- 
dreds. What  is  2  tens  3  units  x  10?  2  hundreds  3  tens  or  230. 
What  is  the  rule  for  multiplying  by  10  ?  Move  each  figure  one  place 
to  the  left." 

110.  Multiplier  Several  Orders.  Multiply  by  the  digit 
in  each  order  separately,  placing  the  right-hand  figure  of 
each  partial  product  under  that  digit  of  the  multiplier 
which  produces  it.  If  one  or  both  factors  are  multiples 
of  10,  multiply  without  regard  to  the  ciphers  and  annex 
the  disregarded  ciphers  to  the  product. 

To  prove,  go  over  the  work  a  second  time.  To  prove 
by  9's,  multiply  the  excess  of  9's  in  the  multiplicand  by 
the  excess  of  9's  in  the  multiplier.  The  excess  of  9's  in  the 
result  should  be  the  excess  of  9's  in  the  original  product. 
Proof  by  9's  is  recommended  from  the  sixth  year  on. 


3768 
386 
22608 
30144 
11304 
1454448 


43000 
52600 
258 


215 


2261800000 


Jill  LESSON   18.    MULTIPLICATION  81 

ILL.   Multiplier  Two  Orders.     Development  exercise.     See  §  66. 

Object  and  Scope.     To  teach  the  multiplication  of  261  by  24. 

Steps.     See  §  10. 

Knowledge.  How  to  multiply  by  a  number  of  one  order  and  how 
to  multiply  by  10. 

Means.  Ask  the  pupils  to  multiply  264  by  24  and  see  what  they 
will  do.  Show  that  24  is  4  +  20,  that  they  may  find  4  264's  and  20 
264's,  and  add  the  results.  They  know  how  to  find  4  264's;  have 
them  do  it.  Show  that  to  find  20  264's  they  must  multiply  by  2  and 
at  the  same  time  move  each  figure  one  place  to  the  left.  Have  them 
finish  the  work.  Have  them  prove  the  work  by  going  over  it  again. 

111.  Mental  Multiplication.     Pupils  should  be  able  to 
call  all  products  to  100  instantly  and  to  multiply  by  a 
number  of  one  order  mentally. 

ILL.  Products  to  100.  2  x  11  =  22,  2  x  12  =  24, .  . .  2  x  49  = 
98,  2  x  50  =  100 ;  3  x  11  =  33,  .  . .  3  x  33  =  99 ;  4  x  11  =  44,  ... 
4  x  25  =  100 ;  5  x  11  =  55,  ...  5  x  19  =  95,  5  x  20  =  100 ;  ... 

ILL.  General  Case.  368  x  8.  Say  2400;  480,  2880;  64,  2944. 
85.37  x  6.  Say,  830;  81-80,  831.80;  42?,  $ 32.22. 

ILL.    Twenty-Nine,  Etc.    39?  x  7.     Say,  $2.80,  7?,  $  2.73. 

ILL.   Aliquot  Parts.     732  x  25.    Multiply  by  100  and  divide  by  4. 

112.  Written  Problems.     See  §  43. 

113.  Exercises.    1.     Teach  the  combinations  of  6's  in  multiplica- 
tion—  development     exercise.      2.  Drill     exercise.      3.  Oral      test. 
4.  Explain  a  written  test  on  the  45  combinations  to  measure  effici- 
ency.   6.  Try  the  exercises  of  §  107.     How  proficient  are  you  ?    6.  In 
the  example  of  §  110  do  you  say  from  habit  in  multiplying  by  6,  48, 
40,  46,  22  ?    If  you  use  more  words  it  will  pay  you  to  change  your 
habit.     7.  Discover  inductively  the  rule  for  the  proof  by  9's.     8.  Ex- 
plain the  proof  by  9's  in  the  examples  of  §  110.     9.  The  commutative 
law  of  multiplication  is,  The  product  of  two  numbers  is  the  same 
whichever  number  is  the  multiplier.     Prove  the  law  (take  3  rows  of 
4  dots  each). 


LESSON  19.  •  DIVISION 

114.  Needs.     The  need  of  division  has  been  developed 
in  §  104.     It  manifests  itself  in  two  phases,  to  find  the 
number  of  equal  parts  (quotition),  and  to  find  one  of  the 
equal  parts    (partition).       Each   phase   is   expressed   in 
several  forms. 

Quotition  Partition 

6?  =  2?  multiplied  by  what?  6?  =  what  multiplied  by  3? 

What  =  6^  divided  by  2?  1  What  =  6?  divided  by  3  ? 

How  many  times  does  6^  con-  What  =  1  of  the  three  equal 
tain2j<?  parts  of  6?!? 

How  many  times  is   2^   con-       What  =  |of6^? 
tained  in  6  j*  ? 

In  quotition,  both  dividend  and  divisor  must  be  of  the 
same  denomination ;  in  partition,  the  divisor  must  always 
be  abstract. 

115.  Means.      The   primary  method   is   counting.      A 
shorter  method  is  subtraction.     The  improved  method  is 
to  form  partial  dividends,  to  divide  each  partial  dividend, 
and  to  add  the  quotients. 

The  improved  method  necessitates  the  mastery  of  the 
combinations  in  division  whose  dividends  are  the  products 
of  the  45  combinations  in  multiplication  and  whose  divisors 
are  the  factors  of  the  combinations  in  multiplication. 
They  are  90  in  number. 

ILL.  In  the  division  of  364  by  7,  the  partial  dividends  are  36  tens 
and  14  units.  The  division  of  36  tens  by  7  necessitates  knowing 
35  -r-  7 ;  the  division  of  14  units  by  7  necessitates  knowing  14  -*-  7. 

82 


§116  LESSON   19.    DIVISION  83 

116.  The  Combinations.  The  combinations  in  division 
are  known  as  soon  as  the  combinations  in  multiplication 
are  known,  but  drill  is  necessary  to  accustom  the  mind  to 
the  new  phase.  The  pupil  is  unsatisfactory  until  he  can 
call  two  results  a  second. 

ILL.  Combinations  of  4's  in  Division.  Drill  exercise.  T.  "We  are 
going  to  drill  on  the  combinations  of  4's  in  division.  Draw  a  circle; 
on  the  outside  write  the  products  of  the  combinations  in  multiplica- 
tion ;  at  the  center  write  4. 


o,  4M.5  4)20;  4)4;  4)_28_;  4)16j 

36 


28 

"John,  read  from  the  circle  in  this  way,  8  4's  are  32,  6  4's  are 
24,  and  so  on.  Jane,  read  in  this  way,  32  contains  4  8  times,  24 
contains  4  6  times,  and  so  on.  You  missed  36  contains  4.  Give  the 
multiplication  table  of  4's.  How  many  times  does  36  contain  4? 
Mary,  read  in  this  way,  4  is  contained  in  32  8  times,  4  is  contained  in 
24  6  times,  and  so  on.  Joseph,  read  in  this  way,  32  -r-  4  =  8, 
24  -4-  4  =  6,  and  so  on.  Henry,  read  in  this  way,  8,  6,  2,  and  so  on. 
As  I  tap  on  the  board,  Peter  may  call  the  results,  8,  6,  2,  and  so  on  in 
perfect  time  (he  taps  at  the  rate  of  two  a  second).  Susan,  read  the 
quotients  at  the  right,  9,  5,  and  so  on  in  perfect  time  (he  taps  at  the 
rate  of  two  a  second). 

"  Nellie,  read  from  the  circle  in  this  way,  ^  of  32  is  8,  {  of  24  is  6, 
and  so  on. 

"  William,  give  me  the  answer  without  explanation.  If  a  boy  rides 
4  mi.  an  hour,  in  how  many  hours  will  he  ride  28  mi.  ?  36  mi.  ?  12  mi.  ? 
Frances,  If  a  boy  rides  36  mi.  in  4  hr.,  how  many  miles  does  he  ride 
in  1  hr.  ?  If  he  rides  20  mi.  in  4  hr.,  how  many  miles  in  1  hr.  ?" 

117.  Partial  Dividends.  First  Form.  The  improved 
method  necessitates  finding  quotients  when  the  dividends 
are  not  multiples  of  the  divisors.  For  a  time  daily  drills, 
and  later  drills  at  intervals  throughout  the  course  are  of 
value.  Charts  like  the  following  are  recommended. 


84 


LESSON   19.    DIVISION 


§118 


They  help  pupils  to  associate  each  dividend  with  the  divi- 
dend which  is  a  multiple  of  the  divisor,  and  they  present 
all  possible  dividends  in  the  same  field. 


DRILL 

FOR  4'8 

DRILL 

FOK 

6'8 

4 

5 

6 

7 

6 

7 

8 

9 

10 

11 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

12 

13 

14 

15 

18 

19 

20 

21 

22 

23 

16 

17 

18 

19 

24 

25 

26 

27 

28 

29 

20 

21 

22 

23 

30 

31 

32 

33 

34 

35 

24 

25 

26 

27 

36 

37 

38 

39 

40 

41 

28 

29 

30 

31 

42 

43 

44 

45 

46 

47 

32 

33 

34 

35 

48 

49 

50 

51 

52 

53 

36 

37 

38 

39 

54 

55 

56 

57 

58 

59 

ILL.  In  the  4's,  the  teacher  points  to  30,  19, 21,  .  .  .  and  the  pupil 
declares  instantly  7  with  2,  4  with  3,  5  with  1,  ... 

Second  Form.  Gratifying  results  will  be  obtained  by 
daily  drills  in  forming  partial  dividends  before  short  di- 
vision is  introduced. 

Take  the  first  or  first  two  digits  for  the  first  partial 
dividend,  use  the  remainder  with  the  next  digit  for  the 
second  partial  dividend,  the  remainder  with  the  next 
digit  for  the  third  partial  dividend,  and  so  on.  Speak  no 
words  except  the  partial  dividends.  Do  no  writing. 

2)978  3)13761  4)27696 

ILL.  1st.  Ex.  Say  9  ;  think  4  with  1  and  say  17 ;  think  8  with  1 
and  say  18 ;  think  9. 

2d.  Ex.  Say  13 ;  think  4  with  1  and  say  17 ;  think  5  with  2  and 
say  26 ;  think  8  with  2  and  say  21 ;  think  7. 

118.  Short  Division.  If  the  preparation  suggested  in 
the  last  paragraph  has  been  made,  short  division  will  be 
performed  as  quickly  as  multiplication  by  a  number  of 
one  order. 


§  119  LESSON  19.    DIVISION  85 

ILL.  T.  "  To-day  we  are  going  to  divide  large  numbers  by  2. 
Divide  973  by  2.  4g6 

2)973  __2 

486  with  1  972 

"  Say  9;  think  4  with  1,  write  4,  and  say  17  :  think  8  with  1,  write 
8,  and  say  13 ;  think  and  write  6  with  1.  How  many  times  does  973 
contain  2?  486  times  with  1  remaining.  How  can  we  prove  the 
answer?  Multiply  486  by  2,  add  1,  and  see  if  the  answer  is  973. 
"  Let  us  see  why  this  process  gives  the  right  result.  The  first  di- 
is  9  groups  of  what?  9  hundreds;  9  hundreds  -4-  2  are  4  hun- 
dreds witji  1  hundred  remaining ;  we  write  the  4  in  hundreds'  order. 
The  1  hundred  remaining  and  the  7  tens  make  17  tens  ;  17  tens  •*•  2 
are  8  tens  with  1  ten  remaining,  and  so  on." 

119.  Long  Division.  Examples  should  be  selected  in 
the  order  of  difficulty  (§  8).  The  bases  of  classification 
are  '  number  of  figures  in  the  divisor,'  and  '  ease  of  find- 
ing quotient  figures.'  Thus,  41  is  an  easier  divisor  than 
212,  or  than  14. 

To  find  the  first  quotient  figure,  divide  the  number  de- 
noted by  the  first  or  first  two  figures  of  the  dividend  by 
the  number  denoted  by  the  first  figure  of  the  divisor. 
Multiply  mentally  the  number  denoted  by  the  first  two 
figures  of  the  divisor  by  the  result.  If  the  quotient  is 
too  large  try  the  next  smaller  and  so  proceed. 

Divisor  Two  Orders.  The  method  of  procedure  is  the 
same  as  in  short  division. 

ILL.  T.  "  To-day  we  are  going  to  divide  by  numbers  of  two 
orders. 

"  Divide  3134  by  41. 


76 


41)3143 

287 

273 

246 

27 


Proof 

76  3116 

41  27 

76  3143 


304 
3116 


86  LESSON   19.    DIVISION  §119 

"  What  is  the  first  partial  dividend  ?  314.  Yes,  because  this  is  the 
least  part  of  the  dividend  that  can  be  divided  by  41.  What  is 
314  -T-  41?  An  easy  way  to  find  the  quotient  is  to  divide  31  by  4. 
The  quotient  appears  to  be  7  but  to  make  sure  that  7  is  not  too  large 
we  will  multiply  41  by  7  mentally.  The  product  is  287  or  less  than 
314.  Where  shall  we  place  7  ?  Yes,  over  4,  the  last  figure  of  the  first 
partial  dividend. 

"How  shall  we  get  the  next  partial  dividend?  Annex  the  next 
figure  of  the  dividend  to  the  remainder.  What  is  the  remainder? 
It  will  be  necessary  to  multiply  41  by  7  and  subtract.  The  next  par- 
tial dividend  is  273.  And  so  on. 

"What  is  3143  -H  41?  76  with  27.  Prove  your  answer.  State  a 
problem  in  which  this  division  is  necessary." 

Divisor  N  Orders.  To  prove,  go  over  the  work  a 
second  time  ;  or  multiply  the  divisor  by  the  quotient  and 
add  the  remainder. 

To  prove  by  9's,  multiply  the  excess  of  9's  of  the  divisor 
by  the  excess  of  9's  of  the  quotient  and  add  the  excess  of 
9's  of  the  remainder.  The  excess  of  9's  of  the  result 
should  be  the  excess  of  9's  of  the  dividend.  Proof  by  9's 
is  recommended  from  the  sixth  year  on. 
ILL.  T.  "  Divide  184570  by  2976. 

62 

2976)184570  6 

17856  8 

6010  48 

5952  _4 

58  7V 

"  What  is  the  first  partial  dividend  ?  18457.  Yes,  because  this  is 
the  least  part  of  the  dividend  that  can  be  divided  by  2976.  Mary, 
explain  finding  the  first  quotient  figure. 

1  18  -r-  2  =  9,  29  x  9  =  261,  9  is  too  large ;  29  x  8  =  232,  8  is  too 
large ;  29  x  7  =  203,  7  is  too  large ;  29  x  6  =  174,  6  is  right.' 

"  Prove  by  9's.  9,  15,  6  (excess  of  the  divisor) ;  8  (excess  of  the 
quotient)  ;  48  (the  product)  ;  13,  4  (excess  of  the  rem.)  ;  52,  7  (excess 
of  the  result)  ;  9,  13,  18,  25,  7  (excess  of  the  dividend)  ;  check." 


§120  LESSON   19.    DIVISION  87 

Divisor  a  Multiple  of  10.  From  the  first,  pupils  should 
be  taught  to  divide  by  a  multiple  of  10  in  such  a  way  as 
to  avoid  the  repetition  of  ciphers. 

Divide  by  the  highest  power  of  10  which  is  a  factor  of 
the  divisor,  and  then  divide  the  quotient  by  the  other 
factor  of  the  divisor.  The  true  remainder  is  the  second 
remainder  multiplied  by  the  power  of  10,  plus  the  first 
remainder. 


ILL.     T.   "  Divide  272568  by  7000. 


Teacher's  Work 


7)272.568, 


38  with  6568 


Mary's  Work 
38 


7000)272568 
21000 


62568 

56000 

6568 


"  Mary's  work  is  right  but  it  is  too  long.  Divide  by  1000  and  then 
by  7.  Dividing  by  1000  the  quotient  is  272  and  the  rem.  568  (make 
a  cross  and  a  point) ;  dividing  272  by  7  the  quotient  is  38  with  a 
rem.  6 ;  the  true  remainder  is  1000  x  6  +  568  or  6568." 

120.  Austrian  Method.  By  the  Austrian  method  the 
product  of  each  digit  of  the  divisor  by  the  quotient  digit 
is  subtracted  as  soon  as  it  is  found  and  the  remainder  alone 
is  written.  It  is  shorter  than  the  common  method  but  is 
not  recommended  because  it  affords  greater  opportunity 
for  error. 

ILL.     T.   "  Divide  184570  by  2976  by  the  Austrian  plan. 

"  The  first  partial  dividend  is  18457  and  the  first  quotient  is  6.  We 
will  multiply  2976  by  6,  subtracting  the  product  of  each  digit  from 
18457  as  we  proceed. 

62 


88  LESSON   19.    DIVISION  §  121 

"  6  6's  are  36,  and  1  are  37,  write  1 ;  6  7's  are  42,  and  3  are  45,  and 
0  are  45,  write  0 ;  6  9's  are  54,  and  4  are  58,  and  6  are  64,  write  6 ;  6 
2's  are  12,  and  6  are  18,  and  0  are  18. 

"  The  next  partial  dividend  is  6010.  2  6's  are  12,  and  8  are  20, 
write  8;  2  7's  are  14,  and  2  are  16,  and  5  are  21,  write  6 ;  2  9's  are  18, 
and  2  are  20,  and  0  are  20 ;  2  2's  are  4,  and  2  are  6,  and  0  are  6." 

121.  Mental  Division.     Pupils  should  be  able  to  call  all 
the  factors  of  numbers  to  100  instantly  and  to  divide  by 
a  number  of  one  order  mentally. 

ILL.     Factors  to  100-     See  §  132. 

ILL.  General  Case.  Proceed  as  in  short  division.  Thus,  439  H-  5. 
43  -T-  5,  8  with  3 ;  39  -f-  5,  7  with  4.  Ans.  87  with  4. 

ILL.   Aliquot  Parts.     475  -T-  25.     Multiply  by  4  and  divide  by  100. 

122.  Sequence  of  Signs.     To  avoid  confusion  it  is  agreed 
that  the  sign  '  x  '  or  '  -5-  '  shall  be  used  before  the  sign 
*  +  '  or   the   sign  '  —  .'     There   is   no   agreement   as   to 
whether  '  x  '  or  '  -s-  '  shall  be  used  first.     The  exact  mean- 
ing should  be  expressed  by  means  of  a'parenthesis.     Some 
writers  employ  these  signs  with  the  understanding  that 
they  are  to  be  used  in  the  order  of  their  occurrence. 

ILL.  T.  "  What  is  the  value  of  6  +  8  x  2  ?  28  if  the  signs  are 
used  as  they  occur,  or  22  if  the  sign  '  x  '  is  used  first.  It  is  agreed  to 
use  <  x  '  or  '  -T-  '  before  '  +  '  or  '  —  .'  22  is  the  right  answer. 

"  What  is  the  value  of  16  -*-  8  x  2  ?  4  if  the  signs  are  used  as  they 
occur,  or  1  if  the  sign  <  x  '  is  used  first.  There  is  no  agreement  here. 
If  such  an  expression  occurs  in  a  test  use  the  signs  as  they  occur. 
The  proper  way  is  to  use  a  parenthesis.  (16  -T-  8)  x  2  =  4 ;  16  -s- 
(8x2)  =  1." 

123.  Operations   Combined.     Pupils    enjoy  greatly   the 
four  operations  combined  within  the  limits  of  the  multi- 
plication table.     A  comma  should  be  written  after  each 
number  to  prevent  confusion. 

ILL.  T.  "  Give  me  the  final  result  instantly.  Write  no  figure.  6,  x 
8,  -f  1,  H-  7,  x  8,  +  9,  -4-  5,  —  1,  -r-  4.  (As  the  teacher  says  '  6  multi- 


§  124  LESSON   19.    DIVISION  89 

plied  by  8,'  the  child  thinks  48;  as  he  says  'plus  1,'  the  pupil  thinks 
49 ;  and  so  on)." 

124.  Explanation  of  Problems.     See  Lessons  6',  7,  and  8. 

125.  Written  Problems.     See  Lesson  9. 

126.  Exercises.     1.     Read:    3x8^;8^x3;  explain   how  these 
readings  come  about.     2.    Set  an  oral  test  on  the  combinations  of  7 
in  division.     Take  the  test  and  determine  your  own  efficiency.    3.  Set 
a  written  test  on  the  90  combinations  in  division.     Take  the  test  and 
determine  your  own  efficiency.     4.   What  is  the  objection  to  the  ex- 
pression, '  6  goes  into  12  2  times '  ?    5.   Make  a  chart  for  7's  for  the 
first  form  of  drill  in  finding  partial  dividends.     6.    Drill  the  class  on 
the   second  form.     7.    Teach  short  division    by   7.     8.    Teach   long 
division  by  a  divisor  of  four  orders.     9.   Prove  inductively  the  rule 
for  proof  by  9's. 


LESSON  20.     FACTORING 

127.  Need.     The  product  of  two  or  more  numbers  is 
called  a  multiple  of  each  number,  and   each   number   is 
called   a  factor   of   the   product.     The   need   of   finding 
multiples  arises  in  the  preparation  of   fractions   for   ad- 
dition and   subtraction  ;  the  need  of   factoring  arises  in 
the  reduction  of  fractions  to  their  lowest  terms  and  in 
cancellation.     Factoring  is  the  life  of  Arithmetic. 

128.  Composites  and  Primes.     Every  composite  number 
is  made   up  of  prime  numbers.     It  is  worth   while   for 
pupils  to  grasp  this  thought  quite  early. 

ILL.  Classification.  T.  "  Let  us  classify  numbers  with  reference 
to  their  factors." 

1,  2,  3,  f,  5,  ft  7,  &  ?,  JJJ,  11,  tf. 

"  Write  the  numbers  through  12.  What  are  all  the  factors  of 
3?  1  and  3.  What  are  all  the  factors  of  4?  1,  2,  and  4.  What 
difference  do  you  notice?  4  has  other  factors  besides  itself  and  1; 
3  has  not.  Cross  every  number  that  has  other  factors  besides  itself 
and  1 ;  they  are  composite  numbers ;  the  others  are  prime  numbers. 
What  is  a  composite  number?  A  prime  number?  State  all  the 
prime  numbers  to  12." 

129.  Finding  Primes.     Pupils  should  know  the  primes 
through  100  in  connection  with  mental  work  in  multiplica- 
tion and  division.     They  find  the  sieve  method  taught  by 
Eratosthenes  more  than  2000  years  ago  both  interesting 
and  valuable. 

ILL.  T.  "  To-day  we  are  going  to  sift  composites  from  primes. 
You  may  all  write  the  numbers  from  2  through  31  in  rows  of  six 
numbers  each." 

90 


§  130  LESSON  20.    FACTORING  91 

2  3  4  5  0  7 

0        y     ;0     11      ;?     13 
tf       /*?     ^     17     ;?      19 

20  21"         22         2S         24         23 

20  2/        2$        29         $0         31 

"  How  shall  we  find  the  higher  multiples  of  2  ?  By  crossing  every 
2d  no.  after  2;  do  so.  To  find  the  higher  multiples  of  3  cross  every 
3d  no.  after  3.  To  hit  the  multiples  of  4  will  it  be  necessary  to  cross 
every  4th  no.  after  4  ?  No,  because  every  multiple  of  4  is  a  multiple  of 
2  and  has  been  already  crossed.  Cross  the  higher  multiples  of  5.  To 
hit  the  multiples  of  6  will  it  be  necessary  to  cross  every  6th  no.  after 
6?  No,  because  they  are  multiples  of  2  and  have  been  already 
crossed ;  it  is  never  necessary  to  count  by  a  composite  number.  Cross 
all  the  multiples  of  7.  7  x  7  is  49.  If  any  no.  less  than  49  is  divided 
by  7  what  will  be  the  quotient  as  compared  with  7  ?  Less  than  7. 
Yes,  and  we  have  now  crossed  all  the  composites  because  we  have 
counted  by  every  number  less  than  7.  Name  all  the  primes  to  30." 

"  Who  can  tell  how  to  separate  composites  from  primes  ?.  Write 
the  numbers  from  2.  Cross  every  2d  no.  after  2,  every  3d  no.  after 
3,  and  so  on,  counting  by  primes  only  until  a  prime  has  been  used 
whose  product  when  multiplied  by  itself  is  more  than  the  largest  num- 
ber written.  By  the  sieve  method,  find  all  the  primes  to  50." 

Law  of  Primes.  The  association  of  primes  with  multi- 
ples of  6  assists  in  making  their  acquaintance. 

ILL.  T.  "  What  seems  to  be  true  of  prime  numbers  with  reference 
to  multiples  of  6?  5  is  one  less  than  6,  7  is  one  more  than  6 ;  11  is 
one  less  than  12,  13  is  one  more  than  12 ;  every  prime,  no.  except  2 
and  3  seems  to  be  one  less  or  one  more  than  a  multiple  of  6.  Is  it 
also  true  that  every  number  which  is  one  less  or  one  more  than  a 
multiple  of  6  is  a  prime  number  ?  " 

130.  Finding  Prime  Factors.  In  reducing  fractions  to 
their  lowest  terms  and  in  finding  the  least  common  multi- 
ple of  large  numbers,  it  becomes  necessary  to  find  the  prime 
factors  of  a  number.  The  subject  should  be  taught  after 
the  addition  and  subtraction  of  fractions  in  ordinary  use. 


92  LESSON   20.    FACTORING  §  131 

Test  the  divisibility  of  the  number  by  the  successive 
prime  numbers  beginning  with  2  until  that  prime  has  been 
tried  whose  product  when  multiplied  by  itself  is  greater 
than  the  number.  If  a  factor  is  found,  treat  the  quotient 
in  the  same  way,  and  so  proceed.  If  no  factor  is  found, 
the  number  is  a  prime. 

ILL.     T.   "  We  are  going  to  find  the  prime  factors  of  numbers. 

3 1 231 

7[77 
11 

"How  shall  we  find  the  prime  factors  of  231?  What  is  the 
smallest  prime  number?  2.  Is  it  a  factor  of  231?  No.  What  is 
the  next?  3.  Is  it  a  factor  of  231  ?  Yes.  Divide  231  by  3.  What 
is  the  first  prime  number  which  is  a  factor  of  77?  7.  Express  231 
by  its  prime  factors.  231  =  3x7x11. 

"  Find  the  prime  factors  of  101.  101  is  not  a  multiple  of  2,  nor  3, 
nor  5,  nor  7,  nor  11.  Is  it  necessary  to  try  any  prime  no.  greater  than 
11?  No,  because  11  x  11  is  121,  a  number  greater  than  101.  101  is 
a  prime  number." 

131.  Rules  for  Divisibility.     Pupils  should  develop  in- 
ductively the  rules  for  divisibility  and  then  memorize  them. 
They  will  have  occasion  to  use  them  constantly.     See  §16. 

A  number  is  a  multiple  of  2  if  its  last  digit  is  a  multiple  of  2 ;  of  3 
if  the  sum  of  its  digits  is  a  multiple  of  3 ;  of  5  if  the  last  digit  is  a 
multiple  of  5;  of  11  if  the  difference  between  the  sum  of  its  digits  in 
the  odd  orders  and  the  sum  of  its  digits  in  the  even  orders  is  a  mul- 
tiple of  11.  There  is  no  serviceable  rule  for  7. 

A  number  is  a  multiple  of  4  if  the  number  denoted  by  its  last  two 
digits  is  a  multiple  of  4 ;  of  8  if  the  number  denoted  by  its  last 
three  digits  is  a  multiple  of  8 ;  of  9  if  the  sum  of  its  digits  is  a  mul- 
tiple of  9. 

132.  Through  a  Hundred.     Pupils  should  be  drilled  on 
naming   the   factors  of   numbers   to  a  hundred   because 
people  of  all  occupations  use  these  numbers  daily. 


§  133  LESSON   20.    FACTORING  93 

ILL.  T.  "  Call  the  numbers  from  100  backwards  and  give  the  fac- 
tors of  each  in  pairs.  100,  2  and  50,  4  and  25,  5  and  20,  10  and  10 ; 
99,  3  and  33,  9  and  11 ;  98,  2  and  49,  7  and  14 ;  97,  prime ;  and  so  on." 

133.  Greatest  Common  Divisor.     There  is  little  occasion 
in  arithmetic  for  finding  G.  C.  D.     The  subject  should  be 
presented  in  the  later  grades  in  connection  with  finding 
the  L.  C.  M.  bv  factoring.     See  5  135. 

«/  o  o 

134.  Least  Common  Multiple.     The  L.  C.  M.  of  numbers 
must  be  found  as  a  preparation  for  the  addition  and  sub- 
traction of  fractions.     The  work  should  be  done  even  in 
the   lower   grades  by  inspection.      The   occasion   rarely 
arises  for  the  use  of  fractions  whose  least  common  denom- 
inator cannot  readily  be  found  by  this  method.     To  pro- 
vide for  the  unusual,  finding  the  L.  C.  M.  of  large  numbers 
should  be  taught  in  the  advanced  grades  by  the  method 
of  factoring. 

By  Inspection.  To  find  the  L.  C.  M.  of  two  numbers 
by  inspection,  compare  the  successive  multiples  of  the 
larger  with  the  smaller  until  a  multiple  of  the  smaller  is 
found.  To  find  the  least  common  multiple  of  more  than 
two  numbers,  find  the  least  common  multiple  of  two  of 
them,  the  least  common  multiple  of  the  result  and  a  third, 
and  so  on. 

ILL.  T.  "We  are  going  to  find  the  least  common  multiple  of 
numbers.  What  is  the  L.  C.  M.  of  two  or  more  numbers?  The  least 
number  that  will  contain  each  of  them  without  a  remainder.  Find 
the  L.  C.  M.  of  6  and  8.  Yes,  24  is  right.  How  did  you  get  it,  Mary  ? 
The  L.  C.  M.  is  the  least  multiple  of  8  that  is  also  a  multiple  of  6; 
8  is  not  a  multiple  of  6,  16  is  not,  24  is.  Mary  compared  the  multiples 
of  8  with  6.  Would  it  be  right  to  compare  the  multiples  of  6  with  8? 
Yes,  but  more  comparisons  would  be  necessary ;  6  is  not  a  multiple 
of  8,  12  is  not,  18  is  not,  but  24  is.  What  is  the  rule  for  finding  the 
L.  C.  M.  of  two  numbers  ? 


94  LESSON  20.    FACTORING  §  135 

"  What  is  the  L.  C.  M.  of  6,  8,  and.9  ?  The  L.  C.  M.  of  6  and  8  is  24. 
The  L.  C.  M.  of  6,  8,  and  9  must  be  the  L.  C.  M.  of  9  and  24.  24  is  not 
a  multiple  of  9,  48  is  not,  but  72  is.  The  L.  C.  M.  of  6,  8,  and  9  is  72. 
What  is  the  rule  for  finding  the  L.  C.  M.  of  more  than  two  numbers  ?  " 

135.  L.  C.  M.  and  G.  C.  D.  of  Large  Numbers.  It  is 
necessary  to  provide  for  those  rare  cases  of  finding  the 
L.  C.  M.  and  G.  C.  D.  which  cannot  be  treated  by  inspec- 
tion. The  factoring  method  appeals  most  strongly  to  the 
intelligence.  A  skillful  arrangement  of  the  prime  factors 
is  of  service. 

Write  the  numbers  in  a  vertical  line.  Find  the  prime 
factors  of  the  first  number.  Discover  whether  the  prime 
factor  in  the  1st  column  is  a  factor  of  the  2d  number;  if 
so  write  it  in  the  column  of  the  1st  prime  factor  and  write 
the  quotient  as  scratch  work.  Discover  whether  the 
prime  factor  in  the  2d  column  is  a  factor  of  the  quotient ; 
and  so  on. 

To  find  the  L.  C.  M.  take  a  factor  from  every  column. 
To  find  the  G.  C.  D.  take  a  factor  from  every  complete 
column. 

ILL.  T.  "  Sometimes  numbers  are  so  large  that  it  is  not  easy  to 
find  their  L.  C.  M.  and  G.  C.  D.  by  inspection.  How  shall  we  find 
these  properties  of  42,  28,  and  154?  We  will  find  the  prime  factors 
of  each  number,  arranging  them  in  such  a  way  that  each  factor  not 
found  in  a  preceding  column  shall  have  a  column  of  its  own.  Write 
the  numbers  in  a  vertical  line. 


42  =  2  x  3  x  7 
28  =  2         x  7  x  2 
154  =  2         x  7        x  11 


L.  C.  M.  =  2  x  3  x  7  x  2  x  11  =  924 
G.  C.  D.  =  2  x  7  =  14 


Scratch 

14 
2 

77 
11 


"  42  =  2  x  3  x  7.     The  2  of  the  1st  column  is  a  factor  of  28 ;  write 
it  in  the  1st  column  and  write  its  quotient,  14,  at  the  side.    3  of  the 


§136  LESSON  20.    FACTORING  95 

2d  column  is  not  a  factor  of  14.  7  of  the  3d  column  is  a  factor  of  14 ; 
write  it  in  the  3d  column ;  its  quotient,  2,  is  a  prime  number ;  write 
it  in  the  4th  column. 

"  2  of  the  1st  column  is  a  factor  of  154 ;  write  it  in  the  1st  column 
and  write  its  quotient,  77,  at  one  side,  and  so  on  as  before. 

"  The  L.  C.  M.  of  42,  28  and  154  must  contain  every  prime  factor 
of  each  number  or  must  be  made  up  of  a  factor  from  every  column. 

"  The  G.  C.  D.  of  42,  28  and  154  must  contain  every  prime  factor 
that  is  common  to  all  or  must  be  made  up  of  a  factor  from  every 
complete  column." 

136.  Fractions  to  Lowest  Terms.     Divide  both  terms  by 
a  common  factor,  divide  both  terms  of  the  result  by  a  com- 
mon factor,  and  so  proceed.     When  no  common  factor  can 
be  found  by  inspection,  find  the  prime  factors  of  one  of 
the  terms,  and  test  the  other  term  by  these  factors  one 
by  one. 

2541  =  231 
5753     523 

ILL.  By  inspection,  11  is  found  to  be  a  common  factor  of  2541  and 
5753.  By  inspection,  no  common  factor  of  231  and  523  is  found. 
231  =3x7x11;  if  231  and  523  have  a  common  factor  it  must  be  3, 
or  7,  or  11 ;  3  is  not  a  factor  of  523,  7  is  not,  11  is  not.  The  fraction 
is  in  its  lowest  terms. 

137.  Exercises.     1.   Find  the  prime  factors  of  529  and  write  the 
trial  divisors  which  you  use.     2.   Is  211  a  prime  number?     Write  the 
trial  divisors  which  you  use.    3.   Discover  inductively  a  rule  for  the 
divisibility  of  a  number  by  11.   4.  Try  to  discover  inductively  a  rule  for 
the  divisibility  of  a  number  by  7.     What  troubles?    6.   Why  should 
pupils  not  be  taught  to  find  L.  C.  M.  by  dividing  by  any  prime  num- 
ber that  is  a  factor  of  two  of  them,  etc.  ?    6.   Why  should  pupils  not 
be  taught  to  find  G.  C.  D.  by  dividing  the  greater  by  the  smaller,  etc. 
(the  Euclidean  method)  ?    7.   In  reducing  fractions  to  their  lowest 
terms,  is  it  necessary  to  find  the  G.  C.  D.  of  their  terms? 


LESSON  21.    FRACTIONS 

138.  What  Part.  The  need  arises  of  measuring  a  part 
in  terms  of  the  whole.  It  is  satisfied  by  selecting  as  a 
standard  one  of  the  equal  portions  of  the  whole  and  by 
stating  how  many  of  the  standard  make  the  part.  As 
in  other  measurements  (§  9)  the  expression  is  abbreviated 
by  naming  the  standard.  The  result  is  a  fraction,  the 
number  of  the  standard  in  the  part  is  the  numerator,  the 
number  of  the  standard  in  the  whole  is  the  denominator. 
A  fraction  is  written  by  placing  the  numerator  over  the 
denominator  and  by  separating  the  terms  by  a  line. 

ILL.     A  is  what  part  of  B  ? 


A  common  measure  of  A  and  B  is  one  of  the  4  equal  parts  of  B. 
A  is  3  of  the  4  equal  parts  of  B.  Let  us  call  '  one  of  4  equal  parts '  a 
fourth ;  A  is  3  fourths  of  B. 

139.  Teaching  Fractions.  A  concept  is  more  vivid  than 
its  name.  Thus,  '  one  of  four  equal  parts '  is  more  vivid 
than  'a  fourth.'  This  fact  should  be  kept  well  in  mind 
by  the  teacher. 

ILL.   T.  "  We  are  going  to  learn  how  to  express  a  part  of  a  whole. . 


"  Fold  a  piece  of  paper  into  two  equal  parts  and  shade  one  of  them. 
What  part  of  the  paper  is  shaded  ?    One  of  the  two  equal  parts  of  the 

96 


§  140  LESSON  21.    FRACTIONS  97 

whole.  One  of  two  equal  parts  we  call  a  half.  The  shaded  part  is 
one  half  of  the  whole.  One  of  three  equal  parts  is  a  third.  What  is 
one  of  four  equal  parts  ?  Yes,  a  fourth  or  a  quarter.  One  of  five 
equal  parts?  One  of  six  equal  parts? 

"  Fold  a  piece  of  paper  into  4  equal  parts  and  shade  3  of  them. 
What  portion  is  shaded  ?  3  of  the  4  equal  parts  or  3  fourths.  Write 
this,  f.  What  does  the  3  show?  The  no.  of  equal  parts  in  the 
shaded  portion.  It  is  called  the  numerator.  What  does  the  4  show  ? 
The  no.  of  equal  parts  in  the  whole.  It  is  called  the  denominator. 
What  is  the  numerator?  The  denominator?" 

140.  Comparisons.      The    conception    of    fractions    is 
strengthened   by   comparisons.      Representation  of   frac- 
tions  by   one-line   diagrams   as   in    Lesson   3  should   be 
taught  before  reductions  and  the  operations. 

ILL.   See  ILL.  §  12. 

141.  Cardinal    Principles.     The   cardinal  principles  of 
fractions   are  :  multiplying   the    numerator,   multiplying 
the  denominator  ;  dividing  the  numerator,  dividing   the 
denominator  ;  multiplying  both  numerator  and  denomi- 
nator by  the  same  number,  dividing  both  numerator  and 
denominator  by  the  same   number.     They  can  easily  be 
taught  in  two  lessons  by  aid  of   the  following  diagram. 
The  mnemonic  rules   on  p.  20  should   be   called   for   at 
intervals. 

-       «- 


ILL.  See  ILL.  §  20.  The  teacher  spends  one  lesson  on  the  prin- 
ciples involving  multiplication  and  one  lesson  on  the  principles  in- 
volving division,  and  drills  upon  all  the  principles  at  intervals. 

142.  Reductions.  Fractions  are  reduced  to  lowest 
terms  in  order  to  grasp  their  values  as  clearly  as  possible. 


98  LESSON  21.    FRACTIONS  §143 

They  are  reduced  to  higher  terms  as  a  preparation  for  ad- 
dition and  subtraction.  Each  form  of  reduction  should 
be  taught  as  the  need  arises.  The  reductions  of  fractions 
with  large  denominators  should  not  be  presented  until 
pupils  have  mastered  the  operations  upon  fractions  within 
the  field  of  the  multiplication  tables,  and  until  they  have 
been  well  grounded  in  factoring. 

ILL.     To  Higher  Terms.     T.  "  The  other  day  we  found  the  sum  of 
I  and  |  by  diagram.     Let  us  find  the  sum  again  and  study  the  work. 

f 


"  We  find  that  f  +  \  -  T9j  +  TV  What  is  the  first  thing  to  do  in 
adding  these  fractions?  To  reduce  them  to  equivalent  fractions  hav- 
ing their  least  common  denominator.  This  denominator  will  be  the 
least  multiple  of  4  that  is  also  a  multiple  of  3.  4  is  not  a  multiple  of 
3,  8  is  not,  but  12  is.  How  shall  we  get  a  fraction  equal  to  f  whose 
denominator  is  12  ?  Divide  12  by  4  and  multiply  both  terms  by  the 
quotient.  What  right  have  we  to  do  this  ?  Multiplying  both  terms 
of  a  fraction  by  the  same  number  does  not  change  the  value  of  the 
fraction.  In  the  same  way,  reduce  f  to  12ths.  Who  can  tell  me  how 
to  reduce  fractions  to  equivalent  fractions  having  their  least  common 
denominator  ?  " 

ILL.  To  Lower  Terms.  T.  "In  the  addition  of  T\  and  ^  we  ob- 
tain ^f .  Let  us  see  if  we  can  find  a  .simpler  expression. 

"  What  does  ^f  mean  ?  12  of  16  equal  parts.  Draw  a  diagram 
and  see  if  you  can  find  a  simpler  expression.  John  is  right,  f  or  6  of 
8  equal  parts  is  simpler.  Why  is  it  that  -J-|  =  |  ?  Dividing  both 
numerator  and  denominator  of  a  fraction  by  the  same  number  does 
not  change  the  value  of  the  fraction.  Is  there  a  simpler  expression 
than  f  ?  Yes,  f .  Who  can  give  me  a  rule  for  reducing  a  fraction  to 
its  lowest  terms  ?  " 

ILL.     Large  Denominators.     See  §  136. 

143.  Addition  and  Subtraction.  Only  the  addition  and 
subtraction  of  fractions  whose  least  common  denominators 


§144  LESSON   21.    FRACTIONS  99 

can  be  found  by  inspection  should  be  taught  in  the  lower 
grades.  These  operations  upon  fractions  with  large  de- 
nominators may  be  taught  in  the  higher  grades  to  impart  a 
feeling  of  power. 

The  following  forms  are  suggested.  The  L.  C.  D.  is 
written  below  the  main  line  with  a  short  line  above. 
This  marks  the  name  of  the  column.  It  is  then  unneces- 
sary to  write  the  denominator  of  each  fraction  separately. 
After  a  time,  pupils  should  do  this  work  without  writing 
anything  except  the  answers. 

Prove  every  answer. 

For  finding  L.  C.  D.  by  inspection,  see  §  134.  See  also 
ILL.  4,  §  72. 


ADDITION 

SUBTRACTION 

12 

6|    8 

8 

6  | 

8 

2|    9 

9 

2| 

20 

9 

Q  5       17 
yTS     T2 

TZ 

311 

H 

ILL.  Subtraction.  Austrian  Method.  To  2  and  9  12lhs  add  enough 
to  make  a  number  ending  in  8  ISlhs,  or  to  9  12lhs  enough  to  make 

I  unit  and  8  12ihs  or  20  12ihs;  add  11  12ths.     In  practice,  say  8  12ths 
and  12  12ths  are  20  IZths,  9  12ihs  and  11   IZths  are  20  12ihs  (write 

II  Igths)  ;  3  and  3  are  6  (write  3).     See  §  98. 

ILL.  Large  Denominators.  Turn  to  §  135.  Suppose  the  denomi- 
nators are  42,  28,  and  154.  The  L.  C.  D.  is  924.  To  find  the  quo- 
tient of  924  divided  by  each  denominator  do  not  perform  the  division, 
but  take  a  factor  from  each  column  not  represented  by  the  divisor. 
Thus,  924  -f-  42  =  2  x  11  or  22 ;  924  -f-  28  =  3  x  11  or  33  ;  924  -=-  154 
=  3  x  2  or  6. 

144.  Multiplication  and  Division.  In  G-eneral.  The 
teaching  of  the  multiplication  and  division  of  fractions 


100 


LESSON   21.    FRACTIONS 


§144 


has  been  illustrated  on  pp.  20  and  21.  The  abbreviation 
afforded  by  cancellation  ehould  be  clearly  taught.  Two 
principles  are  involved.  The  product  of  two  numbers  is 
the  same  whichever  is  used  as  the  multiplier  (commuta- 
tive law),  and  dividing  both  terms  of  a  fraction  by  the 
same  number  does  not  change  the  value  of  the  fraction. 
Prove  every  answer. 

ILL.  T.  "  You  know  the  rule  for  multiplying  fractions.  Multiply 
the  numerators  for  a  new  numerator  and  the  denominators  for  a  new 
denominator.  We  are  going  to  learn  how  to  shorten  the  process 
when  there  is  a  common  factor  in  a  numerator  and  a  denominator. 

"Multiply  ^  by  f.  Yes,  -ffo  or  ?%.  Instead  of  dividing  both 
terms  of  ^  by  5  after  multiplying,  try  the  effect  of  dividing  5  and 
25  by  5  before  multiplying.  The  result  is  the  same.  Why?  Since 
the  numerators  are  to  be  multiplied,  the  9  and  5  may  be  regarded  as 
changing  places,  ^  x  f .  Then  we  may  divide  both  terms  of  ^  by  5. 
How  shall  we  modify  our  rule  for  multiplying  fractions?  Add  to  it 
canceling  when  possible.  Give  me  the  complete  rule." 

Mixed  Numbers.  If  only  one  of  the  terms  is  a  mixed 
number,  pupils  even  in  the  lower  grades  should  perform 
the  operation  without  reducing  the  mixed  number  to  an 
improper  fraction. 


MULTIPLICATION 

DIVISION 

368| 

7)2580| 

7 

368| 

2580f 

461 

2|        1383 

2|)  1267| 

345f 

11)5071 

922 

461 

1267| 

ILL.   T.    "Multiply  368f  by  7.     Write  as  on  the  board. 
7  times  |  are  4|  (write  f),  60  (write  0),  etc. 


Say, 


§144 


LESSON  21.    FRACTIONS 


101 


"  Divide  2580f  by  7.  Write  as  on  the  board.  Say  25,  think  3 
with  4  and  say  48  ;  think  6  with  6  and  say  60 ;  think  8  with  4  and  say 
4?  or  i^;  think  §  (V  -  7)  and  write  f! 

3          3 

"  Multiply  461  by  2f.  Write  as  on  the  board.  We  nmst  multiply 
by  f  and  then  by  2.  To  multiply  by  f,  we  multiply  by  3  placing  the 
product  to  one  side,  and  then  divide  the  result  by  4.  Do  not  write 
any  more  figures  than  are  on  the  board. 

"Divide  1267f  by  2|.  Write  as  on  the  board.  Multiply  both 
dividend  and  divisor  by  4  to  make  the  divisor  an  integer.  Do  not 
write  any  more  figures  than  are  on  the  board." 

Mixed  Numbers.  If  both  of  the  terms  are  mixed  num- 
bers, pupils  in  the  upper  grades  may  perform  the  opera- 
tions without  reducing  the  mixed  numbers  to  improper 
fractions.  In  division,  after  the  divisor  has  been  changed 
to  an  integer,  pupils  in  the  upper  grades  should  practice 
dividing  by  using  the  factors  of  the  divisor. 


MULTIPLICATION 

DIVISION 

37f           74 

Mi          s* 

18f)704f 

i 

24| 

181 

296 

56)2114 
8)2114 
7)264} 

37 

37f 

704f 

ILL.  T.  "  f  x  \  =  i  ;  37  x  f  =  24f  ;  £  x  18  =  13* ;  37  x  8  =  296 ; 
:57  x  10  =  370." 

ILL.  Div.  T.  "  Those  who  sit  at  my  right  may  divide  2114  by  56  ; 
the  others  may  divide  by  8  and  then  divide  the  result  by  7.  Which 
is  the  easier  process  ?  " 

Remainders.  The  quotient  is  usually  declared  as  a 
mixed  number.  To  express  it  as  a  whole  number  with  a 


102  LESSON   21.    FRACTIONS  §  145 

remainder,  for  the  whole  number  take  the  whole  number 
of  the  mixed  quotient ;  for  the  remainder,  take  the  product 
of  the  divisor  by  the  fraction  of  the  mixed  quotient. 

ILL.  T.  "  How  many  lengths  of  5|  yd.  each  may  be  cut  from  a 
rope  23  yd.  long  and  how  long  will  be  the  remainder  ? 

"  Mary's  answer  is  4  lengths  and  2  yd.  Prove  it.  4  lengths  are 
22  yd. ;  22  yd.  +  2  yd.  are  24  yd.  Something  is  wrong.  23  -=-  5£  = 
4^-;  there  are  4^  lengths,  but  not  4  lengths  and  2  yd.  The  re- 
mainder is yT  of  a  length  or  ^  of  *£  yd.  or  1  yd." 

145.  Simple  and  Complex.     The  terms  of  a  fraction  may 
both  be  integers  or  not  both  integers.     This  gives  rise  to 
simple  fractions  and  complex  fractions.     A  fraction  of  a 
fraction  is  called  a  compound  fraction. 

To  simplify  complex  fractions,  simplify  each  term  sepa- 
rately, and  divide  the  numerator  by  the  denominator. 
This  subject  should  be  omitted  from  elementary  schools. 

146.  Mental   Work.     Many  drills  should  be   given   in 
performing  the  operations  mentally.     Multiplication  and 
division  demand  special  attention. 

ILL.  T.  "  How  shall  we  multiply  a  fraction  by  an  integer  men- 
tally ?  Multiply  the  numerator  or  divide  the  denominator.  4  x  ^  ? 
|£  (multiply  the  nu.).  4  x  ^\?  f  (divide  the  den.). 

"  How  shall  we  divide  a  fraction  by  an  integer  mentally  ?  Divide 
the  numerator  or  multiply  the  denominator,  f  -H  2  ?  f  (divide  the 
nu.).  |  -4-  2  ?  T\  (multiply  the  den.)." 

147.  Problems.     See  Part  III. 

148.  Exercises.     1.   Set  an  oral  test  for  the  sixth  year  on  the  four 
operations  with  fractions.     2.    A  written  test.     3.    Give  a  first  lesson 
on  the  addition  of  fractions.     4.    On  the  multiplication  of  fractions. 
5.    On  the  division  of  fractions.     6.    How  many  times  can  8|  gal.  be 
taken  from  100  gal.  and  how  many  gallons  will  remain  ?    7.    Prove 
the  answer.     8.    Explain  as  to  a  class. 


LESSON  22.     DECIMALS 

149.  Need  and  Means.  In  measuring  a  part  in  terms  of 
the  whole  the  need  arises  of  a  uniform  set  of  standards  in 
place  of  half,  third,  fourth,  and  so  on.  This  need  is  satis- 
fied by  extending  the  decimal  plan  of  expressing  how  many. 
Just  as  one  of  the  10  equal  parts  of  a  hundred  is  ten,  and 
one  of  the  10  equal  parts  of  ten  is  a  unit,  in  like  manner 
one  of  the  10  equal  parts  of  a  unit  is  a  tenth,  one  of  the  10 
equal  parts  of  a  tenth  is  a  hundredth,  and  so  on.  It  only 
remains  to  use  a  decimal  point  to  mark  where  units  end 
and  tenths  begin  in  order  to  express  both  integers  and 
fractions  by  the  same  plan. 

A  decimal  is  a  fraction  expressed  in  tenths,  hundredths, 
thousandths,  and  so  on,  by  aid  of  a  point. 

ILL.  T.  "  We  are  going  to  find  a  new  way  of  expressing  a  part. 
Let  us  divide  a  whole  into  10  equal  parts,  each  small  part  into  10  equal 
parts,  and  so  on. 

"  What  is  one  of  the  10  equal  parts  of  a  whole  ?  A  tenth.  One  of 
the  10  equal  parts  of  a  tenth?  A  hundredth.  One  of  the  10  equal 
parts  of  a  hundredth  ?  A  thousandth.  You  may  draw  a  diagram  like 
mine  on  the  board.  A  is  what  part  of  £?  7  tenths  5  hundredths 
or  75  hundredths. 


B 

438.75 

"  Let  us  find  a  shorter  way  of  writing  this.  You  may  write  438 
and  place  after  it  a  period,  which  we  will  call  a  decimal  point,  to  show 
that  8  is  in  units'  order,  and  after  that  you  may  write  seven  five. 

103 


104  LESSON  22.    DECIMALS  §  150 

"  What  is  one  of  the  10  equal  parts  of  a  hundred  ?  Ten.  One  of 
the  10  equal  parts  of  ten  ?  A  unit.  One  of  the  10  equal  parts  of  a 
unit?  A  tenth.  Then,  in  what  order  is  the  7?  Tenths.  What  is  one 
of  the  10  equal  parts  of  a  tenth?  A  hundredth.  Then,  in  what  order 
is  the  5  ?  Hundredths.  Read  438.75.  438  and  7  tenths  5  hundredths 
or  438  and  75  hundredths. 

"Write:  6  tenths,  .6;  4  tenths,  4\  3  tenths  2  hundredths,  .82; 
4  hundredths,  .04,  Read :  .2,  2  tenths ;  .03,  8  hundredths ;  .27,  27  hun- 
dredths or  2  tenths  7  hundredths." 

150.  Reading.    Pupils  should  be  taught  to  read  decimals 
by  calling  the  name  of  each  standard  and  also  by  calling 
the  name  of  the  lowest  standard.     They  should  memorize 
the  names  of  the  first  six  standards  by  number  and  find 
the  names  of  lower  standards  by  numerating. 

ILL.  T.  "  Read  .638  calling  the  name  of  each  order.  6  tenths  3  hun- 
dredths 8  thousandths.  Read  calling  the  name  of  the  lowest  order  only. 
638  thousandths. 

"Commonly,  we  read  the  name  of  the  lowest  order  only.  I  want 
you  to  memorize  the  numbers  of  the  orders  by  name.  Thus,  the  1st 
is  tenths ;  the  2d,  hundredths;  the  3d,  thousandths;  the  fourth,  ten- 
thousandths;  the  fifth,  hundred -thousandths;  the  sixth,  millionths. 
Read  .00007,  7  hundred-thousandths,  because  7  is  in  the  fifth  order. 
Read  .00000758;  758  hundred-millionths,  because  the  6th  order  is 
millionths,  the  next  ten-millionths,  and  the  next  hundred-millionths." 

151.  Writing.     Pupils  should  be  taught  to  write  the 
number  of  the  lowest  standard,  and  to  express  the  name 
of  the  standard  by  reviving  in  memory  the  names  of  the 
first  six    standards   and  by  finding  the  names  of   other 
standards  by  numerating. 

ILL.  T.  "  Write  6  thousandths;  .006.  Yes,  write  6,  the  number  of 
thousandths,  remember  that  thousandths  is  the  3d  order  and  place  the 
decimal  point  so  as  to  show  3  decimal  places.  Write  6  ten-millionths. 
Write  6,  the  number  of  ten-millionths,  remember  that  millionths  is 
the  6th  order,  find  that  ten-millionths  must  be  the  7th,  and  place  the 
decimal  point  so  as  to  show  7  decimal  places." 


§  152 


LESSON  22.    DECIMALS 


105 


152.  Addition  and  Subtraction.  Since  decimals  are 
written  by  the  same  plan  as  integers,  their  addition  and 
subtraction  may  be  performed  by  the  same  plan. 

Prove  every  answer. 


ADDITION 

SUBTRACTION 

.97 

2.06 

.438 

.087 

1.408 

1.973 

ILL.  T.  "  In  writing  integers  for  addition  and  subtraction,  where 
did  we  place  digits  of  the  same  order?  In  the  same  column.  We 
must  do  the  same  with  decimals.  Be  sure  to  place  the  decimal 
points  in  the  same  column." 

153.  Multiplication  and  Division.  Since  decimals  are 
written  by  the  same  plan  as  integers,  their  multiplication 
and  division  may  be  performed  by  the  same  plan.  Special 
attention  must  be  paid  to  the  decimal  points. 


MULTIPLICATION 

5.375             2 
.68             5 

DIVISION 

.68 

2 
5 

5X375.)3X655.00 
3  2250 

43000 
32250 

43000 
43000 

10 

3.65500             lv 

Multiplication  by  an  Integer.  Multiply  as  in  integers 
and  point  off  as  many  decimal  places  in  the  product  as 
there  are  decimal  places  in  the  multiplicand. 

ILL.  T.  "  Multiply  .28  by  4.  Nellie  gets  112.  This  cannot  be 
right  because  .28  is  less  than  1  and  the  product  must  be  less  than 
1x4.  She  has  neglected  the  decimal  point.  The  answer  should  be 
112  hundredths  just  as  28  apples  x  4  is  112  apples.  112  hundredths  is 
1.12.  There  must  be  the  same  number  of  decimal  places  in  the 


106  LESSON  22.    DECIMALS  §  154 

product  as  in  the  multiplicand.     Who  will  give  me  the  rule  for  mul- 
tiplying a  decimal  by  an  integer?" 

Multiplication  by  a  Decimal.  Multiply  as  in  integers  and 
point  off  as  many  decimal  places  in  the  product  as  there 
are  decimal  places  in  both  multiplicand  and  multiplier. 

ILL.     First  Lesson.     See  p.  17. 

ILL.  Later  Lesson.  "  Be  sure  to  prove  your  answer  by  going  over 
the  work  again.  Check  the  decimal  point  by  finding  the  product  of 
the  first  significant  digit  in  each  term.  In  the  above,  5  x  .6  =  3. 
The  decimal  point  is  in  the  right  place." 

Division  by  an  Integer.  Divide  as  in  integers  and  point 
off  as  many  decimal  places  in  the  quotient  as  there  are 
decimal  places  in  the  dividend. 

ILL.     See  Multiplication  by  an  Integer  above. 

Division  by  a  Decimal.  Move  the  decimal  point  in  both 
dividend  and  divisor  so  as  to  make  the  divisor  an  integer 
and  proceed  as  in  the  division  of  a  decimal  by  an  integer. 

ILL.  T.  "Divide  .76  by  A.  Mary  gets  .19.  This  cannot  be  right 
because  .76  -4-  4  =  .19.  She  has  neglected  the  decimal  point  of  the 

divisor. 

Teacher's  Work  Mary's  Work 

.4)  .76 


1.9 


.19 


"How  shall  we  proceed?  We  will  make  the  divisor  an  integer. 
How  can  we  do  this?  By  moving  the  decimal  point  one  place  to  the 
right  in  both  dividend  and  divisor.  We  may  do  this  because  multi- 
plying both  dividend  and  divisor  by  the  same  number  does  not 
change  the  value  of  the  quotient.  Cross  out  the  old  points  and  write 
the  new.  Finish  the  work.  What  is  the  rule  for  dividing  a  decimal 
by  a  decimal  ?  " 

•  154.  Reductions.  It  would  be  possible  to  dispense  with 
fractions  other  than  decimals  just  as  it  would  be  possible 
to  dispense  with  systems  of  weights  and  measures  other 
than  the  decimal,  but  as  long  as  both  common  fractions 


§  154  LESSON   22.    DECIMALS  107 

and  decimals  are  used  the  need  will  arise  of  changing 
from  one  form  to  the  other. 

In  the  lower  grades  these  reductions  should  be  made 
with  those  fractions  only  which  can  be  reduced  to  deci- 
mals exactly,  viz.,  with  fractions  whose  denominators 
have  no  other  prime  factors  than  2  and  5. 

In  the  higher  grades,  the  reduction  of  other  fractions 
may  be  introduced.  It  is  then  necessary  to  show  clearly 
how  results  are  expressed,  both  exactly  and  approxi- 
mately, and  that  a  common  fraction  appended  to  a  deci- 
mal cannot  occupy  a  decimal  place  but  must  be  of  the 
denomination  of  the  figure  which  it  immediately  follows  : 


COM.  FKAC.  TO  DEO. 

7)3.0000 

.4285| 

f  =  .4$  ex. ;  .4  approx. 
7  =  .42f  ex. ;   .43  approx. 


DEC.  TO  COM.  FKAO. 

.4f  =  ^  -  10  =  f 

.42f  =  yp-  H-  100  = 

.428|  =  a-°/a  -*-  1000 


ILL.  Lower  grades.  T.  "  After  a  part  has  been  expressed  by  a 
common  fraction  it  is  sometimes  necessary  to  express  it  by  a  decimal 
and  vice  versa. 

"Change  f  to  a  decimal,  £  means  3-^-4.  The  easiest  way  is  to 
perform  the  division.  Do  so.  £  =  .75. 

"  Change  .75  to  a  common  fraction.  .75  means  Jfa.  Reduce  it  to 
its  lowest  terms.  ^  =  f." 

ILL.  Upper  grades.  T.  "  Reduce  f  to  a  decimal.  Does  the  divi- 
sion terminate  ?  No.  If  there  is  any  prime  factor  in  the  denomi- 
nator which  is  not  found  in  10  the  division  can  never  end.  What 
shall  we  do?  To  get  the  exact  result,  we  may  stop  dividing  at  any 
point  and  write  after  the  last  figure  of  the  quotient  the  fraction  whose 
numerator  is  the  last  remainder  and  whose  denominator  is  the 
divisor.  To  get  an  approximate  result,  we  may  stop  dividing  at  any 
point,  increasing  the  last  figure  of  the  quotient  by  1  if  the  next  fig- 
ure of  the  quotient  would  be  5  or  more. 


108  LESSON   22.    DECIMALS  §155 

"  Let  us  examine  our  first  result,  ,4f .  Is  the  j  tenths,  or  hun- 
dredths?  We  divided  3  by  7  or  30  tenths  by  7.  Hence,  our  result 
must  be  4f-  tenths  just  as  30  apples  divided  by  7  is  4$  apples.  Re- 
member that  a  common  fraction  can  never  occupy  a  decimal  order 
but  is  of  the  denomination  of  the  figure  which  it  immediately 
follows." 

The  teacher  takes  the  pupil  into  his  confidence  and  discusses  all 
that  is  suggested  in  the  examples  worked  above. 

155.  Degree  of   Approximation.     The  nature  of  every 
problem   suggests    the    number   of    decimal    places   that 
should  be  used  in  its  solution.     The  attention  of  pupils  in 
the  later  grades  should  be  called  to  this  matter. 

ILL.  T.  "  You  should  always  consider  the  number  of  decimal 
places  that  must  be  used  in  a  computation. 

"  We  find  the  interest  of  $  1  to  be  $  .185£.  To  find  the  interest  of 
$2596,  to  what  place  should  £  be  reduced?  We  want  the  answer 
true  to  cents ;  we  agree  to  count  5  mills  or  more  as  1^.  Hence,  we 
must  carry  the  answer  to  the  3d  decimal  place.  The  highest  order 
of  2596  is  the  3d  place  to  the  left  of  units;  the  3d  to  the  left  of 
units  must  be  multiplied  by  the  6th  to  the  right  of  units  to  make  the 
3d  to  the  right  of  units.  The  multiplicand  must  be  $.185833. 

"We  find  from  the  tables  that  the  amount  of  $1  is  $1.6958814, 
what  is  the  amount  of  $6.75.?  Reasoning  as  before,  we  find  that 
$  1.695  may  be  multiplied  by  6.75." 

156.  Use  of  Factors.     It  is  often  of  advantage  to  multi- 
ply or  to  divide  by  the  factors  of  a  number  instead  of  by 
the  number  itself.     This  device  is  of  special  value  in  divi- 
sion where  the  quotient  is  desired  without  consideration 
of  the  remainder. 


132.7098-48 

272568  -=-  7000 

8)132.7098 
6)16.5887+ 
2.7647+ 

7)272.568X 
38.938+ 
See  P.  87. 

§  157  LESSON   22.    DECIMALS  109 

157.  Mental  Work.     The   decimal   equivalents   of    the 
business  fractions  should  be  memorized  and  applied  : 

1.1     2.13.1     2     34.1     5.1357 
2" '    ^'  ^  '    I'  4  '    "5"'  "5"'  "5"'   5  '    6"'  6" '    IP  $»  "g"'   8  * 

158.  Problems.     See  Part  in. 

159.  History.     Decimals  were  invented  in  the  sixteenth 
century,  but  the  decimal  point  was  not  introduced  until 
the  next  century.     For  a  hundred  years,  the  number  of  a 
decimal   order  was  expressed   by  an  Arabic  or   Roman 
numeral  above  or  to  the  right  of  the  order. 

0123  O  I  II  III 

5967;  5(0)9(1)6(2)7(3);  5967;  5.967 

160.  Exercises.     1.   Explain  the  plan  of  writing  dollars,  cents,  and 
mills.     2.   Read  400.008  and  .408.     3.   Give  the  rule  for  the  use  of 
and  in  reading  decimals.     4.   Add  .4§  aud  .86.     6.   In  Xo.  4,  why  can- 
not 6  be  added  to  f,  making  the  answer  1.26|?    6.   Multiply  14.2f 
by  .82^  and  get  the  exact  answer.    7.   In  No.  6,  express  the  multipli- 
cand and  multiplier  without  the  use  of  common  fractions  with  just 
enough  decimal  places  in  each  to  get  the  answer  true  to  two  decimal 
places.     Three  decimal  places  are  to  be  found  in  the  answer,  but  only 
two  are  to  be  retained. 


LESSON  23.     DENOMINATE  NUMBERS 

161.  How  Much.     The  need  of  expressing  how  much 
arises  as  early  as  the  need  of  expressing  how  many.     It  is 
satisfied  by  measuring  as  explained  in  §  9.     The  results  of 
the  measurements  are  denominate  numbers. 

The  relative  place  of  denominate  numbers  appears  in 
the  following  classification. 

A  number  may  have  its  unit  expressed  or  not  expressed.  This 
gives  rise  to  concrete  numbers  and  abstract  numbers.  Concrete,  2  gal- 
lons, 2  men;  abstract,  2. 

A  concrete  number  may  have  its  unit  in  the  tables  expressing  how 
much  or  may  not  have  its  unit  there.  This  gives  rise  to  denominate 
numbers  and  to  not-denominate  numbers.  Denominate  number, 
2  gallons. 

A  denominate  number  may  have  its  value  expressed  by  one  unit  or 
by  more  than  one  unit.  This  gives  rise  to  simple  denominate  num- 
bers and  compound  denominate  numbers.  Simple  den.  no.,  2  gallons ; 
compound  den.  no.,  2  gallons  3  quarts. 

162.  Teaching.     Teachers  should  develop  the  tables  in 
accordance  with  the  laws  of  measurement  in  such  a  way 
as  to  excite  interest,  and  should  then  insist  upon  their 
thorough   mastery.     The   old   plan   was   to   require   the 
memorizing  of  the  tables  with  no  explanations.     Let  us 
not  err  in  the  other  direction  by  giving  developments  and 
requiring  no  memorizing. 

ILL.   Long  Measure.     See  III.  §  9. 

ILL.  Square  Measure.  T.  "How  large  is  this  blackboard?  It  is 
8  ft.  long  and  4  ft.  wide.  Yes,  that  is  a  good  way  of  expressing  it. 
I  am  going  to  teach  you  another  way. 

110 


§162  LESSON  23.    DENOMINATE  NUMBERS  111 

"  Here  is  a  piece  of  paper  1  ft.  long  and  1  ft.  wide.  We  call  its 
area  a  square  foot.  Let  us  find  how  many  such  pieces  will  cover  the 
board.  Yes,  we  might  put  this  piece  on  the  board  as  many  times  as 
necessary  and  count  the  times,  but  there  is  a  better  way.  If  the  board 
is  8  ft.  long  how  many  such  pieces  would  make  1  strip?  If  the  board 
is  4  ft.  wide  how  many  such  strips  would  there  be  ?  What  is  the  area 
of  the  board?  32  square  feet. 

"  Suppose  we  want  to  express  the  area  in  square  inches.  Find  how 
many  square  inches  make  a  square  foot.  How  did  you  get  144,  Mary  ? 
'  There  are  12  sq.  in.  in  1  strip  and  12  strips.'  Excellent." 

He  continues  in  a  similar  way  with  the  other  units  and  insists  upon 
the  memorizing  of  the  table. 

ILL.  Time.  T.  "  How  many  days  are  there  in  a  week  ?  What  do 
you  mean  by  a  day?  The  time  from  sunrise  to  sunrise.  Yes,  or  the 
time  it  takes  the  earth  to  make  one  rotation  on  its  axis  (he  explains 
the  phenomenon). 

"  How  old  are  you,  John  ?  What  do  you  mean  by  a  year  ?  We 
really  mean  the  time  it  takes  the  earth  to  make  one  revolution  about 
the  sun  (he  explains  the  phenomenon). 

"How  many  days  make  a  year?  How  do  you  suppose  this  was 
found  out  in  the  first  place  ?  The  shadow  of  a  pole  was  measured 
every  day  and  the  number  of  sunrises  counted  from  the  time  the 
shadow  was  the  shortest  until  it  was  the  same  length  again.  We 
could  test  the  length  of  a  year  in  the  same  way,  but  we  have 
better  methods.  The  ancients  found  only  360  days,  but  we  know 
now  that  there  are  365  days  and  a  fraction  of  a  day  Q  da.  less  674 
seconds)." 

In  a  similar  way,  he  creates  interest  in  each  of  the  other  units. 

ILL.  Weight.  T.  "  How  much  do  you  weigh,  Henry?  What  do 
you  mean  by  a  pound  ?  The  pound  used  by  the  grocer  is  the  weight 
of  7000  grains  of  wheat.  Here  is  a  bag  of  wheat.  I  am  going  to  ask 
you  to  count  out  7000  grains  (he  distributes  the  wheat).  As  soon  as 
any  one  gets  100  grains  he  may  bring  them  to  Henry,  who  will  put 
them  into  a  bag.  John  may  make  a  mark  on  the  board  for  every 
hundred.  How  many  hundreds  shall  we  need  ?  At  last  I  have  7000 
grains  of  wheat.  Here  are  scales.  Joseph,  you  may  weigh  the  wheat 
and  see  how  the  weight  agrees  with  the  count."  And  so  on. 


112  LESSON  23.    DENOMINATE  NUMBERS  §162 

Neglected  Weights.  Apothecaries'  weight  and  apothe- 
caries' fluid  weight  are  excluded  from  some  courses  of 
study.  It  seems  a  pity  because  they  are  used  in  the  house- 
hold constantly.  They  can  be  made  both  helpful  and 
interesting. 

ILL.  T.  "  What  is  the  least  amount  of  a  fluid  as  of  water  ?  Yes,  a 
drop.  How  many  drops  of  water  make  a  teaspoonful  ?  To-night  you 
may  find  out. 

"  How  many  drops  did  you  find,  John  ?  100.  Joseph  ?  120.  I 
think  the  number  is  from  100  to  120.  The  druggists  count  60.  Who 
will  find  out  and  tell  us  to-morrow  how  many  teaspoonfuls  of  water 
weigh  an  ounce  ?  Henry,  we  shall  depend  on  you. 

"  How  many  did  you  find,  Henry  ?  6.  The  druggists  count  8. 
How  many  ounces  of  water  make  a  pint  ?  You  may  find  out  later. 
There  is  an  old  saying  that  I  want  you  to  remember,  '  A  pint's  a 
pound  the  world  round.'  This  means  a  pint  of  water  weighs  a  pound. 
It  is  nearly  correct.  16  oz.  of  water  make  a  pound.  Write  the  table 
of  Apothecaries'  fluid  weight.  60  drops  make  a  spoonful,  8  spoon- 
fuls make  an  ounce,  16  ounces  make  a  pint,  8  pints  make  a  gallon." 

Instruction  by  Problems.  In  the  upper  grades  important 
facts  may  be  brought  out  by  a  series  of  problems. 

ILL.   T.  "  Let  us  find  a  rule  for  correcting  the  calendar. 

"  A  year  lacks  674  sec.  of  365£  days.  What  objection  is  there  to 
counting  365  da.  to  the  year?  After  a  time  summer  would  come  in 
the  dead  of  winter.  What  is  the  correction  for  \  da.  ?  One  year  out 
of  every  4  is  counted  as  a  leap  year  with  366  da.  Every  year  that  is  a 
multiple  of  4  is  a  leap  year. 

"  This  plan  makes  every  year  674  sec.  too  long.  How  many  days 
of  error  are  there  in  100  yr. ?  Ans.  f  da.  nearly.  How  is  this  error 
corrected  ?  Out  of  every  400  years,  three  years  that  would  be  counted 
as  leap  years  by  the  first  plan  are  counted  as  common  years.  To  be  a 
leap  year  every  century  year  must  be  a  multiple  of  400. 

"  By  the  last  plan  how  many  seconds  of  error  are  there  in  100 
years?  Ans.  2600.  In  how  many  years  will  there  be  a  day's  error? 

s.  3323. 

"  Memorize  these  rules." 


§163 


LESSON  23.    DENOMINATE  NUMBERS 


113 


163.  Reductions.  Every  exercise  in  reducing  from 
higher  to  lower  denominations  or  from  lower  to  higher 
denominations  should  be  treated  as  a  problem.  The 
teacher  should  keep  in  mind  that  very  little  of  this  work 
is  done  outside  of  the  schoolroom. 


930  in.,  ?  integers  of  h.  den. 

930  in.  =  77  ft.  6  in. 

77  ft.  =  25  yd.  2  ft. 

25  yd.  =  4  rd.  3  yd. 
930  in.  =  4  rd.  3  yd.  2  ft.  6  in. 
Ans. 


4  rd.  3  yd.  2  ft.  6  in.,  ?  inches 

4  rd.  3  yd.  =  25  yd. 

25  yd.  2  ft  =  77  ft. 

77  ft.  6  in.  =  930  in.     Ans. 


EXPL.  How  many  feet  in  930  in.?  77  ft.  6  in.  How  many  yards 
in  77  ft.  ?  25  yd.  2  ft.  How  many  rods  in  25  yd.  ?  4  rd.  3  yd. 

For  the  remainder  in  25  yd.  -4-  5£  yd.  see  §  144,  remainders. 

EXPL.  How  many  yards  in  4  rd.  3  yd.  ?  25.  How  many  feet  in 
25  yd.  2  ft.  ?  77.  How  many  inches  in  77  ft.  6  in.  ?  930. 

164.  Operations.  The  operations  should  be  treated 
exactly  as  the  operations  upon  integers.  Attention  is 
called  to  increasing  a  given  date  by  a  number  of  days  and 
to  finding  the  number  of  days  from  one  date  to  another  as 
is  necessary  in  problems  in  interest. 


June  29  +  90  da.,?  date 

June  29  to  Sept  27,  ?  days 

June  119 

June,  1 

July  89 

July,  31 

Aug.  58 

Aug.,  31 

Sept.  27  Ans. 

Sept.,  27 

Days,  90  Ans. 

EXPL.  June  29  +  90  da.  is  June  119  or  July  89  (June  has  30 
da.)  .  .  . 

EXPL.  In  June  there  is  one  day  left;  in  July  there  are  31 
days;  .  .  . 


114  LESSON   23.     DENOMINATE  NUMBERS  §  165 

165.  Mental  Work.     Nearly  all  of  the  work  should  be 
mental.     See  §  70. 

166.  The  Metric  System.     The  need  arises  of  a  uniform 
system  of  weights  and  measures.     It  is  satisfied  by  the 
decimal  plan  ;  ten  units  of  one  denomination  make  one  of 
the  next  higher.     The  names  of  the  prefixes  below  units 
are  taken  from  the  Latin  and  the  names  of  the  prefixes 
above  units  are  taken  from  the  Greek  ;  they  are  abbrevi- 
ated by  their  initial  letters,  small  for  the  Latin  and  capi- 
tal for  the  Greek.     The  units  are  determined  systematic- 
ally from  the  distance  from  the  equator  to  the  north  pole  ; 
they   are   abbreviated    by   their   initials.     No   period   is 
placed  after  an  abbreviation. 

The  metric  system  bears  the  same  relation  to  English 
weights  and  measures  that  decimals  bear  to  common  frac- 
tions. Each  system  has  its  advantages  ;  they  are  not  all 
in  favor  of  the  metric. 

Teaching  Lower  Grades.  The  nickel  (5  r  piece)  should 
be  made  the  basis  of  teaching  the  metric  system.  Pupils 
should  not  think  in  English  denominations  and  then  give 
the  metric  equivalents,  but  should  proceed  as  if  there  were 
no  English  denominations. 

ILL.  T.  "Yesterday  I  asked  each  of  you  to  bring  a  nickel  to 
class.  From  it  we  are  going  to  learn  a  new  system  of  weights  and 
measures. 

"The  nickel  is  2  millimeters  thick.  How  thick  is  the  cover  of 
your  arithmetic?  How  thick  is  your  penholder?  We  express  very 
short  lengths  in  millimeters.  When  you  study  physics  and  chemis- 
try later  on,  you  will  have  much  to  do  with  this  new  system.  Talk 
with  the  folks  at  home  about  it. 

"Place  5  nickels  in  a  pile.  What  is  the  thickness  of  the  pile? 
10  millimeters  is  called  a  centimeter.  How  long  is  your  little  finger? 
How  long  is  your  penholder  ? 


§  166 


LESSON   23.    DENOMINATE  NUMBERS 


115 


"The  nickel  is  nearly  2  centimeters  in  diameter,  exactly  2.1. 
Place  5  nickels  in  a  row.  The  row  will  be  10.5  centimeters  long. 
Draw  a  line  as  long  as  the  row ;  erase  .5  centimeters  or  5  millimeters 
of  it.  The  line  is  1  decimeter  long.  How  long  is  your  arm  ? 

"  Henry,  draw  on  the  blackboard  a  line  10  decimeters  long  ;  it  is  a 
meter  long.  What  is  the  width  of  this  room  ? 

"  10  meters  make  1  dekameter,  10  dekameters  make  1  hectometer, 
10  hectometers  make  1  kilometer.  Meters  and  kilometers  are  the 
units  in  common  use.  What  is  the  length  of  a  city  block  in  meters? 
About  75.  How  many  blocks  make  a  kilometer  ?  About  12|.  How 
far  is  it  from  10th  St.  to  135th  St.  ?  About  10  kilometers. 

"  Who  will  give  me  the  table  for  long  measure  ?  10  mm  =  1  cm,  etc. 
Does  this  table  remind  you  of  any  other?  10  mills  =  1  cent,  etc." 

The  teacher  continues  the  exercise  and  proceeds  in  a  similar  way 
with  the  measures  of  weight  (the  nickel  weighs  5  grams)  and  of 
capacity.  At  the  end,  he  insists  upon  thorough  memorizing  of  the 
tables. 

Teaching  Higher  Crrades.  A  thorough  mastery  of  the 
metric  system  may  be  gained  by  a  study  of  the  following 
table.  From  the  knowledge  of  how  each  unit  is  obtained 
and  the  value  of  the  meter,  each  of  the  English  equivalents 
may  be  computed. 


TABLE 

UNIT  ' 

AB. 

How  OBTAINED 

ENGLISH  EQUIVALENT 

Length 

meter 

m 

.0000001  distance  from 

39.37  in.  ; 

the  equator  to  the  pole 

1  Km  =  f  mi.  (ap.) 

Weight 

gram 

g 

wt.  1  cu  cm  of  water 

15.432  gr.  ; 

lKg  =  2£lb.  (ap.) 

Capacity 

liter 

1 

1  cu  din 

1  qt.  (ap.) 

Land 

are 

a 

10  m  x  10  m 

A  A.  (ap.) 

Wood 

stere 

8 

Imxlmxlm 

i  cd.  (ap.) 

ILL.  T.  "Kilometer  is  used  in  the  metric  system  where  mile  is 
used  in  the  English.  A  meter  is  39.37  in.  Compute  the  value  of  a 
kilometer  in  terms  of  the  fraction  of  a  mile. 


116 


LESSON  23.  .  DENOMINATE  NUMBERS 


§167 


"Kilogram  is  used  in  the  metric  system  where  pound  is  used  in  the 
English.  A  grarn  is  the  weight  of  1  cu  crn  of  water  ;  a  cubic  foot  of 
water  weighs  62|  Ib.  Compute  the  weight  of  a  kilogram  in  terms  of 
pounds." 

History.  The  metric  system  was  invented  in  France 
under  direction  of  Napoleon  and  is  now  used  by  most 
peoples  except  those  who  speak  English.  In  the  United 
States  it  was  legalized  in  1866  but  its  use  at  present  is 
confined  chiefly  to  scientific  works. 

167-  Exercises.  1.  Teach  cubic  measure.  2.  Which  is  heavier 
and  by  how  many  grains,  a  pound  of  feathers  or  a  pound  of  gold  ? 
3.  An  ounce  of  feathers  or  an  ounce  of  gold?  4.  Was  1900  a  leap 
year  ?  Why  ?  6.  Below  are  given  the  usual  solutions  of  the  prob- 
lems of  §  163.  Criticise  them.  6.  Teach  the  table  of  weight  in  the 
metric  system,  using  a  nickel  as  the  basis.  7.  Solve  the  last  problem 
of  §  166. 


12 1 930  in. 

3177ft.  6  in. 
5£  [25  yd.  2ft. 
4  rd.  3  yd. 
930  in.  =  4  rd.  3  yd.  2  ft.  6  in. 


4  rd.  3  yd.  2  ft.  6  in. 

J| 

25yd. 

_3 

77ft. 
J2 
930  in. 


LESSON  24.     MENSUKATION 

168.  Forms.  In  accordance  with  the  laws  of  logical 
division  pupils  may  be  led  to  create  for  themselves  the 
various  geometrical  forms.  They  will  respond  with  de- 
light and  enthusiasm,  and  will  gain  a  grasp  of  the  subject 
which  will  be  a  constant  joy. 

Major  Forms.  A  subject  is  made  more  vivid  by  con- 
sidering it  from  different  points  of  view. 

ILL.  T.  "  We  may  consider  that  which  has  no  dimension,  that 
which  has  one  dimension,  that  which  has  two  dimensions,  and  that 
which  has  three  dimensions.  This  gives  rise  to  points,  lines,  surfaces, 
and  solids.  This  block  (he  shows  a  cube)  is  a  solid,  its  faces  are 
surfaces,  its  edges  are  lines,  and  its  vertices  are  points. 

"  We  may  consider  points,  lines,  and  surfaces  as  moving  and  leav- 
ing paths.  What  is  the  path  of  a  moving  point  ?  What  is  the  path 
of  a  moving  line  ?  What  is  the  path  of  a  moving  surface  ? 

"  What  is  the  intersection  of  two  plane  surfaces  ?  What  is  the 
intersection  of  two  lines  ?  " 

Lines  and  Angles.  The  teacher  develops  horizontal, 
perpendicular,  vertical,  and  oblique  lines  as  below. 

ILL.  T.  "  A  point  may  move  constantly  in  the  same  direction  or 
constantly  in  a  different  direction.  This  gives  rise  to  straight  lines 
and  curved  lines. 

"  Draw  straight  lines  in  pairs  and  find  what  is  true  of  their  meet- 
ing. They  meet  or  do  not  meet.  This  gives  rise  to  angles  and  to 
parallel  lines.  -» 

"  Draw  lines  in  pairs  that  meet.  What  is  true  of  their  adjacent 
angles?  They  are  equal  or  are  not  equal.  This  gives  rise  to  right- 
angles  and  oblique  angles. 

"Compare  oblique  angles  with  right-angles.  An  oblique  angle  is 

117 


118  LESSON   24.    MENSURATION  §168 

greater  than  a  right  angle  or  less  than  a  right  angle.     This  gives  rise 
to  obtuse  angles  and  acute  angles." 


Dimensions.  Pupils  should  not  only  use  the  word, 
dimension,  but  should  know  what  it  means. 

ILL.  T.  "  Let  us  find  what  is  meant  by  one  dimension,  two 
dimensions,  and  three  dimensions. 

"  Look  at  an  edge  of  this  cube ;  it  has  only  one  dimension.  Look 
at  a  face;  at  a  point  on  the  face  how  many  lines  can  be  drawn  each 
of  which  is  perpendicular  to  each  of  the  others?  Only  two.  The  face 
has  two  dimensions. 

"  Hold  three  pencils  in  such  a  way  that  each  shall  be  perpendicular 
to  each  of  the  others.  Can  you  hold  four  pencils  in  such  a  way? 
Look  at  three  edges  of  a  cube  meeting  at  a  point.  What  is  true  of 
these  edges  ?  Each  is  perpendicular  to  each  of  the  others.  The  cube 
has  three  dimensions  because  at  every  point  within  it  three  lines  may 
be  perpendicular  each  to  each  of  the  others.  Can  you  image  a  form 
that  has  more  than  three  dimensions  ?  " 


Polygons.  Quadrilaterals,  and  triangles  according  to 
their  sides,  have  been  developed  in  §  4  and  §  5. 

ILL.  Triangles.  T.  "Let  us  classify  triangles  according  to  their 
angles. 

"  How  many  right  angles  can  a  triangle  have  ?  (They  determine  by 
experiment.)  How  many  obtuse  angles ?  How  many  acute  angles? 
This  gives  rise  to  right-angle  triangles,  obtuse  angle  triangles,  and 
acute  angle  triangles." 

ILL.  Regular  Polygons.  T.  "  The  sides  and  angles  of  a  polygon 
may  all  be  equal  or  not  all  equal.  This  gives  rise  to  regular  and 
irregular  polygons. 


§168  LESSON  24.    MENSURATION  119 

"  Draw  a  regular  polygon  of  three  sides,  of  four  sides,  of  five  sides, 
of  six  sides,  ...  of  an  infinite  number  of  sides.  This  gives  rise  to 
equilateral  triangles,  squares,  regular  pentagons,  regular  hexagons,  .  .  . 
circles.  Give  me  two  definitions  of  a  circle.  A  circle  is  a  regular 
polygon  of  an  infinite  number  of  sides.  A  circle  is  a  plane  surface 
bounded  by  a  curved  line  every  point  of  which  is  equally  distant  from 
the  center." 


/A    D     O    Q     O\ 

Altitudes.  Pupils  should  understand  clearly  that  a 
triangle  may  have  three  altitudes  —  one  for  each  base. 

T.  "  The  altitude  of  a  triangle  or  quadrilateral  is  a  line  drawn 
from  any  vertex  perpendicular  to  the  opposite  side.  Draw  an  acute 
angle  triangle,  an  obtuse  angle  triangle,  a  parallelogram,  and  a  trape- 
zoid ;  draw  the  altitude  of  each  by  a  dotted  line." 


Pyramids  and  Prisms.  The  common  forms  should  be 
classified. 

ILL.  T.  "  A  solid  may  have  a  polygon  for  its  base  and  equal 
triangles  for  its  faces,  or  a  polygon  for  its  base  and  equal  rectangles 
for  its  faces,  or  various  other  forms,  but  we  shall  consider  the  first 
two  kinds  only.  This  gives  rise  to  pyramids,  or  to  cones  when  the 
base  of  the  pyramid  is  a  circle  ;  and  to  prisms,  or  to  cylinders  when  the 
base  of  the  prism  is  a  circle.  A  pyramid  or  cylinder  is  named  from 
its  base.  Here  are  (he  shows  the  forms)  a  square  pyramid  and  a 
cone,  a  square  prism  and  a  cylinder." 


Regular  Solids.     Attention  should  be  called  to  the  reg- 
ular tetrahedron  and  to  the  cube  as  leading  up  to  the  sphere. 

ILL.     T.  "  A  solid  may  have  all  of  its  faces  equal  polygons.     Here 
are  a  regular  tetrahedron,  a  cube,  and  a  sphere.     The  word,  tetrahe- 


120 


LESSON  24.    MENSURATION 


169 


dron  is  the  Greek  f  or  four-base.  Define  a  regular  tetrahedron  ;  a  cube. 
Give  two  definition  s  for  a  sphere.  A  sphere  is  a  regular  solid  with  an 
infinite  number  of  faces.  A  sphere  is  a  solid  bounded  by  a  curved 
surface,  every  point  of  which  is  equally  distant  from  the  center." 


169.  Constructions.  Pupils  should  be  required  to 
make  solids  from  pasteboard.  The  teacher  who  is  not 
willing  to  direct  this  work  will  do  well  to  omit  solids 
from  the  course. 

ILL.  Prisms  and  Cylinders.  T.  "  We  will  make  a  triangular 
prism.  On  the  pasteboard  upon  your  desk  draw  two  parallel  lines 
4  in.  apart  (he  draws  upon  the  board  as  he  instructs).  Lay  off  3  dis- 
tances of  2  inches  on  each  line ;  connect  the  points  of  division ;  con- 
struct equilateral  triangles.  Prepare  flaps  for  pasting  as  represented 
by  the  dotted  lines.  Cut  entirely  through  outside  lines  and  partly 
through  inside  lines.  Fold  and  paste  the  flaps  on  the  inside. 

"  We  will  make  a  cylinder.  On  the  sheet  of  paper  upon  your  desk 
draw  two  parallel  lines  4  in.  apart.  Lay  off  6  in.  on  each  line ;  con- 
nect the  points  of  division ;  prepare  the  flap.  Cut  entirely  through 
outside  lines.  Roll,  and  paste  the  flap  on  the  inside.  You  do  not 
need  a  base." 


ILL.  Pyramids  and  Cones.  "  We  will  make  a  cone.  Use  the  sheet 
of  paper  upon  your  desk.  From  any  point  as  a  center  with  a  radius 
of  5  in.  draw  an  arc  of  a  circle.  Lay  off  any  convenient  distance,  as 
BC ';  draw  AD  for  the  flap.  Cut  entirely  through  the  outside  lines; 
roll  the  form  until  AB  coincides  with  AC,  and  paste  the  flap  on  the 
inside. 


§  170  LESSON  24.     MENSURATION  121 

"We  will  make  a  square  pyramid.  Proceed  as  in  drawing  the 
form  for  a  cone,  using  the  pasteboard  upon  your  desk,  and  making 
BC  the  same  length  as  before.  Divide  BC  into  4  equal  parts  and 
draw  the  chords.  Connect  the  points  of  division  with  A.  Construct 
a  square  on  one  of  the  sides.  Prepare  flaps  for  pasting  as  represented 
by  dotted  lines.  Cut  entirely  through  outside  lines  and  partly 
through  inside  lines.  Fold,  and  paste  the  flaps  on  the  inside." 


170.  Circumference  of  Circle.     On  p.  13,  the  circumfer- 
ence has  been  found  to  be  3^  x  D  or  trD. 

171.  Areas.     Rectangle.     The  area  of  a  rectangle  is  the 
number  of  square  units  in  one  strip  multiplied   by  the 
number  of  the  strips. 

ILL.    T.     He  proceeds  as  suggested  in  §  162,  ILL. 

Parallelogram.     The  area  of  a  parallelogram  is  the  area 
of  a  rectangle  which  has  the  same  base  and  altitude. 
ILL.     T.  He  proceeds  as  in  §  13,  ILL. 

Triangle.  The  area  of  a  triangle  is  half  the  area  of  a 
rectangle  which  has  the  same  base  and  altitude. 

ILL.  T.  "  Find  by  paper  cutting  (§  14,  Ex.  5)  the  relation  of  a 
triangle  to  a  parallelogram  which  has  the  same  base  and  altitude. 
"What  is  the  relation  of  a  parallelogram  to  the  rectangle  which  has 
the  same  base  and  altitude  ?  What  is  the  area  of  a  triangle  ?  " 

He  shows  the  same  by  paper  folding  (§  14,  Ex.  6). 

Trapezoid.  The  area  of  a  trapezoid  is  half  the  sum  of 
the  areas  of  two  rectangles  one  of  which  has  the  upper 
base  and  the  other  the  lower  base  and  both  the  altitude  of 
the  trapezoid. 

ILL.  T.  "  Into  what  does  the  diagonal  of  a  trapezoid  divide  the 
figure  ?  What  is  the  area  of  a  trapezoid  ?  " 


122  LESSON  24.    MENSURATION  §  172 

Circle.  The  area  of  a  circle  is  the  area  of  a  square 
whose  side  is  the  diameter,  multiplied  by  ^TT.  Pupils 
should  visualize  a  circle  as  inscribed  within  a  square  and 
should  remember  that  the  ratio  of  the  circle  to  the  square 
is  3|  :  4. 

ILL.  T.  "  Inscribe  a  circle  within  a  square.  Divide  the  circle  into 
equal  parts  by  radii.  The  circle  is  the  sum  of  an  infinite  number  of 
small  triangles ;  the  sum  of  the  bases  of  the  triangle  is  the  circum- 
ference of  the  circle,  irD ;  the  altitude  of  the  triangles  is  the  radius, 
-Z).  Hence,  the  area  of  the  circle  is  half  the  area  of  a  rectangle  whose 
base  is  irD  and  whose  altitude  is  \  D,  or  ~  x  -n-D  x  -  D,  or  - 


172.  Convex  Surfaces.  The  convex  surface  of  a  solid  is 
all  the  surface  except  its  bases.  The  forms  of  prisms, 
cylinders,  pyramids,  and  cones,  cut  out  ready  for  pasting, 
suggest  the  rules. 

Prisms.  The  convex  surface  of  a  prism  or  cylinder  is  the 
area  of  a  rectangle  whose  base  is  the  perimeter  of  the  base 
of  the  solid  and  whose  altitude  is  the  altitude  of  the  solid. 

Pyramids.  The  convex  surface  of  a  pyramid  or  cone  is 
half  the  area  of  a  rectangle  whose  base  is  the  perimeter  of 
the  solid  and  whose  altitude  is  the  slant  height  of  the  solid. 

Spheres.  The  surface  of  a  sphere  is  IT  times  the  area  of 
a  square  whose  side  is  the  diameter. 

ILL.   "  The  surface  of  a  sphere  is  4  times  the  area  of  a  circle  which 
has  the  same  diameter  (§  13,  ILL.).     The  area  of  the  circle  is  l- 
Hence,  the  surface  of  the  sphere  is  4  x  ~  irD1*,  or  irD2." 


173.    Volumes.     The  volume  of  a  solid  is  the  number  of 
cubic  units  in  its  contents. 


§  173  LESSON  24.    MENSURATION  123 

Prisms.  The  volume  of  a  prism  or  cylinder  is  the 
number  of  cubic  units  in  one  layer  multiplied  by  the  num- 
ber of  the  layers. 

ILL.  T.  He  follows  the  plan  suggested  for  visualizing  in  §  80,  ILL. 
The  cylinder  is  a  variety  of  the  prism. 

Pyramids.  The  volume  of  a  pyramid  or  cone  is  one  third 
the  volume  of  a  prism  which  has  the  same  base  and  altitude. 

ILL.  T.  "  To  find  the  relation  of  a  pyramid  to  a  prism,  I  am  going 
to  ask  some  one  to  construct  a  square  prism  and  a  square  pyramid 
with  the  same  base  and  altitude,  the  prism  with  one  base  and  the 
pyramid  with  no  base.  Make  the  pyramid  first ;  the  side  of  the  base 
of  the  prism  will  then  be  known.  We  will  compare  their  volumes  by 
filling  them  with  sand.  John,  we  shall  depend  on  you." 

"  These  forms  are  well  made.  John  may  stand  before  the  class 
and  make  the  comparison.  What  do  you  find?  'The  volume  of  the 
pyramid  is  one  third  the  volume  of  a  prism  which  has  the  same  base 
and  altitude?  '  John  has  done  well." 

Spheres.  The  volume  of  a  sphere  is  the  volume  of  a 
cube  whose  edge  is  the  diameter,  multiplied  by  ^  TT.  Pu- 
pils should  visualize  a  sphere  as  inscribed  within  a  cube 
and  should  remember  that  the  ratio  of  the  sphere  to  the 
cube  is  3^  :  6. 

ILL.  Sphere.  T.  "  Inscribe  a  sphere  within  a  cube.  Image  the 
sphere  as  divided  into  equal  pyramids  whose  vertices  are  at  the  center. 
The  sphere  is  the  sum  of  an  infinite  number  of  small  pyramids ;  the 
sum  of  the  bases  of  the  pyramids  is  the  surface  of  the  sphere,  vDz  • 

the  altitude  of  the  pyramids  is  the  radius  of  the  sphere,  -  D.  Hence, 
the  volume  of  the  sphere  is  one  third  the  volume  of  a  prism  whose 
base  is  vD2  and  whose  altitude  is  |  Z>,  or  -0  x  TrD2  x  \  D,  or  i  irDa." 

%  3  • 


124 


LESSON  24.    MENSURATION 


§174 


174.  Similarity.     If  a  form  is  enlarged  and  the  likeness 
is  preserved  in  every  respect  the  original  and  the  result 
are  said  to  be  similar. 

Multiplying  a  linear  part  by  a  number  multiplies  every 
linear  part  by  that  number ;  multiplies  every  surface  part 
by  the  square  of  that  number;  multiplies  every  solid  part 
by  the  cube  of  that  number. 

ILL.     See  §  68. 

175.  Right-Angle  Triangles.     The  right  angle  triangle 
is  the  basis  of  many  problems.     The  square  of  the  hypote- 
nuse is  equal  to  the  sum  of  the  squares  of  the^  other  two 
sides.     The  ancients  determined  this  law  experimentally ; 
pupils  in  the  elementary  schools  must  be  content  to  do 
likewise. 

ILL.  T.  "  Let  us  discover  the  relation  of  the  hypotenuse  of  a  right 
angle  triangle  to  the  other  two  sides. 

"Draw  two  lines  at  right  angles ;  lay  off  4  equal  spaces  on  one  and 
3  equal  spaces  on  the  other ;  find  how  many  spaces  there  are  oil  the 
hypotenuse  by  measurement. 

"  Do  the  same,  making  12  and  5  equal  parts  on  the  lines. 

"  Do  you  find  a  common  relation  between  the  hypotenuse  and  the 
other  sides?  32  +  42  =  52;  52  +  122  =  132.  What  seems  to  be  the 
relation  of  the  hypotenuse  of  a  right-angle  triangle  to  the  other  sides  ? 


176.  Exercises.  1.  Teach  horizontal,  vertical,  and  oblique  lines. 
2.  Draw  an  oblique  triangle  and  indicate  its  three  altitudes  by  dotted 
lines.  3.  Construct  a  3-in.  cube  of  pasteboard.  4.  Construct  a 
regular  tetrahedron  of  pasteboard.  5.  Develop  the  rule  for  convex 
surface  of  a  pyramid.  6.  Teach  the  rule  for  the  volume  of  a  rec- 
tangular prism. 


LESSON    25.     INVOLUTION,    EVOLUTION    AND   LOGA- 

RITION 

177.  Needs.     A  product  may  be  separated  into  a  num- 
ber of  equal  factors.     Thus,  8  =  2  x  2  x  2,  or  8  =  23. 

Let  us  classify  the  needs  which  arise  from  this  state- 
ment by  the  omission  of  each  term  in  succession  (§  7).  If 
the  product  is  wanting,  the  requirement  becomes  No.  1 
and  gives  rise  to  involution  ;  if  the  equal  factor  is  want- 
ing, the  requirement  becomes  No.  2  and  gives  rise  to  evo- 
lution ;  if  the  number  of  times  the  equal  factor  occurs  is 
wanting,  the  requirement  becomes  No.  3  and  gives  rise  to 
logarition.* 

1.  What  =  23  ? 

2.  8  =  what8  ?    or  What  =  v/8? 

3.  8  =  2  vhat  ?    or  What  =  log  82  ? 

178.  Involution.     Involution  is  the  process  of  finding 
the  product  from  the  equal  factor  and  the   number  of 
times  the  factor  occurs.     The  product  is  the  power,  the 
equal  factor  is  the  base,  the  number  of  times  the  factor 
occurs  is  the  index  of  the  power  or  the  exponent. 

ILL.  T.  "  There  is  a  short  way  of  expressing  that  the  same  num- 
ber is  used  several  times  as  a  factor. 

"  Write  8  =  2x2x2.  The  number  of  times  2  is  used  as  a  factor 
is  written  over  and  a  little  to  the  right  of  2.  Thus,  8  =  28.  2  is 
called  the  base  and  3  is  called  the  exponent.  28  is  read  '  2  to  the  3d 
power  '  or  '  the  cube  of  2.'  What  does  3'2  mean  ?  It  is  read  '  3  to  the 
2d  power '  or  '  the  square  of  3.'  How  is  26  read  ?  What  does  it 
mean?  What  is  its  value ?  What  is  2  called?  What  is  5  called? 

*  A  word  suggested  for  the  phrase,  the  process  of  finding  logarithms. 

125 


126  LESSON   25.    INVOLUTION  AND  EVOLUTION  §179 

"  The  process  of  finding  the  product  of  equal  numbers  is  called  invo- 
lution ;  the  word  itself  means  rolled  up.  The  product  is  called  the 
power.  In  what  way  is  involution  performed  ?  By  multiplication." 

Use.  The  principal  use  of  involution  is  to  afford  an 
abbreviated  form  of  expression.  If  a  number  is  to  be 
raised  to  a  high  power,  the  law  for  multiplying  when  the 
bases  are  the  same  may  be  used  to  advantage. 

ILL.  The  amount  of  $1  for  20  yr.  at  6%  compound  interest  is 
<$(1.06)20.  To  make  20  separate  multiplications  would  be  a  long 
process.  The  result  can  be  found  by  5  multiplications.  Thus,  1.06 
x  1.06  =  1.1236  or  (1.06)2;  1.12362  =  1.2625  or  (1.06)4;  1.26252  = 
=  1.5938  or  (1.06)8;  1.59382  =  2.5404  or  (1.06)16;  2.5404  x  1.2625  = 
3.2071  or  (1.06)i6  x  (1.06)4  or  (1.06)20. 

179.  Evolution.  Evolution  is  the  process  of  rinding  the 
equal  factor  from  the  product  and  the  number  of  times 
the  factor  occurs.  The  product  is  not  named,  the  equal 
factor  is  the  root,  the  number  of  times  the  factor  occurs  is 
the  index  of  the  root. 

ILL.  T.  "It  is  sometimes  necessary  to  find  one  of  the  equal  fac- 
tors of  a  number.  Can  you  give  me  an  illustration  ?  If  the  cube  of 
a  number  is  8,  what  is  the  number  ?  Yes,  or  if  the  volume  of  a  cube 
is  8  cu.  in.,  what  is  its  edge? 

"Find  the  number  whose  oth  power  is  32.  How  did  you  get  2, 
Henry?  '  I  found  the  prime  factors  of  32.  32  =  26,  or  2  is  one  of  the 
5  equal  factors  of  32.'  Excellent. 

"  It  is  convenient  to  give  names  to  the  process  and  to  the  terms. 
The  process  is  called  evolution ;  the  word  itself  means  unrolled.  The 
equal  factor  sought  is  called  the  root ;  why  is  root  a  good  name 
for  the  term?  The  number  of  equal  factors  is  called  the  index  of 
the  root.  To  find  the  number  whose  5th  power  is  32  is  to  find  the 
«  5th  root  of  32.'  The  <  fifth  root  of  32  '  is  written  >/32.  The  sign,  ^, 
is  a  modification  of  r,  the  initial  of  root.  When  the  index  is  2  it  is 
not  written." 

The  Process  for  Pupils.  A  root  which  can  be  exactly 
expressed  by  the  decimal  notation  can  be  found  by  factor- 


§179 


LESSON   25.    INVOLUTION  AND  EVOLUTION 


127 


ing.  Every  root  can  be  found  by  trial.  It  is  recom- 
mended that  the  extraction  of  roots  be  limited  in  elemen- 
tary schools  to  these  two  methods. 

ILL.  By  Factoring.  T.  "  Find  the  number  whose  square  is  576  or 
find  the  square  root  of  576.  By  factoring,  we  find  576  =  22  x  22  x  22 
x  32  =  242;  V576  =  24.  Find  ^9261." 

ILL.  By  Trial.  T.  "  Find  V231.  231  =  3  x  7  x  11 ;  V231  cannot 
be  found  by  factoring.  What  shall  we  do  ? 

"  We  will  find  by  trial  two  numbers  differing  by  1  unit,  such  that 
the  square  of  one  shall  be  less  than  231  and  the  square  of  the  other 
greater  than  231.  The  smaller  will  be  V23~I  true  to  units.  152  =  225 ; 
162  =  256 ;  V231  =  15  +. 

"  We  will  find  by  trial  two  numbers  each  of  which  is  '  15  +  '  differ- 
ing by  1  tenth,  such  that  the  square  of  one  shall  be  less  than  231  and 
the  square  of  the  other  greater  than  231.  The  smaller  will  be  \/231 
true  to  tenths.  15.12  =  228.01 ;  15.22  =  231.04 ;  V231  =  15.1  +. 

"  In  a  similar  way  we  can  find  the  answer  true  to  any  required  place. 
Can  the  answer  be  expressed  exactly  by  the  decimal  notation  ? 

"  Find  V2T5  to  one  decimal  place." 


3  9261 

1.3 
1.3 
1.69 
1.3 
507 
169 

1.4 
1.4 
1.96 
1.4 

784 
196 

3  3087 

3  1029 

7|343 
7[49 
7 

2.JL97 

3/ 

2.744 
£5  =  1.3+ 

The  Process  for  Teachers.  To  extract  the  nth  root  of 
an  integer,  point  off  into  periods  of  n  figures  each  begin- 
ning with  units'  place,  extract  the  root  of  the  number  de- 
noted by  the  first  two  periods,  then  the  root  of  the  number 
denoted  by  the  first  three  periods,  and  so  on. 

As  a  guide,  raise  a  +  b  to  the  wth  power  and  proceed  as 
the  formula  indicates. 


128  LESSON   25.    INVOLUTION  AND  EVOLUTION  §179 

ILL.     Extract  the  cube  root  of  1860867. 

(a  +  b)3  =  a3  +  3  a2b  +  3  ab*  +  bs 
=  a3  +  (3  a?  +  3  ab  +  b*)b 


1'860'867|123 


300 
60 


364 


43200 
1080 


44289 


860 


728 


132867 


132867 


a  =  10 


a  =  120 
b  =  S 


Preparation.     As  a  guide  we  raise  a  '+  b  to  the  3d  power  and  factor. 

We  separate  the  number  into  periods  of  three  figures  each  because 
the  cube  root  of  the  number  denoted  by  the  first  period  will  give  the 
first  figure  of  the  root,  the  cube  root  of  the  number  denoted  by  the 
first  two  periods  will  give  the  first  two  figures  of  the  root,  and  so  on. 
(The  teacher  should  satisfy  himself  of  these  facts  inductively.) 

First  Extraction.  We  extract  the  cube  root  of  1'860  to  obtain  the 
first  two  figures  of  the  root. 

If  we  extract  the  cube  root  of  a3,  we  get  a,  the  first  term  of  the 
root  in  the  guide;  hence,  if  we  extract  the  cube  root  of •  1,  we  must 
get  the  first  figure  of  the  root  in  this  example ;  the  cube  root  of  1  is  1, 
or  the  first  figure  is  1 ;  a  =  10  because  1  is  not  1  unit  but  1  thousand. 
Subtracting  the  value  of  a8,  we  get  860  which  equals  3  a?b  +  3  «62  +  b8. 

If  we  divide  3  a2b  by  3  a2,  we  get  Z>,  the  second  term  of  the  root  in 
the  guide ;  hence,  if  we  divide  what  corresponds  to  3  a2b,  or  the  greater 
part  of  860,  by  what  corresponds  to  3  a2,  we  must  get  the  second  figure 
of  the  root  in  this  example ;  3  a2  =  300 ;  860  -*-  300  =  2,  or  b  =  2. 

If  we  multiply  what  is  within  the  parenthesis  by  b,  we  get  the  rest 
of  the  power  in  the  guide ;  hence,  if  we  multiply  what  corresponds  to 
what  is  within  the  parenthesis  by  what  corresponds^  to  b,  we  should 
get  the  rest  of  the  power  in  this  example  if  it  is  a  perfect  power ;  3  a2 
=  300 ;  3  ab  =  60 ;  62  =  4 ;  the  parenthesis  =  364  ;  multiplying  by  b, 
or  2,  and  subtracting,  we  get  132.  Hence,  the  cube  root  of  1'860  is 
12+,  and  the  remainder  found  by  subtracting  the  cube  of  12  from 
1860  is  132. 


LESSON  25.    LOGARITION  129 

Second  Extraction.  We  extract  the  cube  root  of  1860'867  to  get 
the  first  three  figures  of  the  root;  we  may  regard  the  whole  as  1860 
thousands  867  units. 

We  know  that  the  new  a  is  120  because  1860  is  not  1860  units  but 
1860  thousands,  and  that  the  remainder  after  subtracting  the  cube  of 
120  from  1860867  is  132867  which  =  3  a*b  +  3  aW  +  bs.  Hence,  we 
start  in  at  once  to  find  the  new  b. 

If  we  divide  3  a2ft  by  3  a2,  we  get  b,  the  second  term  of  the  root  in 
the  guide ;  hence,  if  we  divide  what  corresponds  to  3  a2b,  or  the 
greater  part  of  132867,  by  what  corresponds  to  3  a2,  we  must  get  the 
next  figure  of  the  root  in  this  example ;  3  a2  =  43200 ;  and  so  on. 

The  Process  for  Teachers.  To  extract  the  root  of  a 
fraction  extract  the  root  of  the  numerator  and  then  the 
root  of  the  denominator.  To  extract  the  root  of  a  deci- 
mal, point  off  into  periods  beginning  with  the  decimal 
point  and  proceed  as  with  integers,  pointing  off  as  many 
decimal  places  in  the  root  as  there  are  decimal  periods  in 
the  decimal. 

ILL.  The  square  root  of  ff  =  £  ;  the  square  root  of  f  =  V.66'66+. 
The  square  root  of  .00365  =  V.00'36'50.  The  teacher  should  discover 
why  the  decimal  is  pointed  off  from  the  left  rather  than  from  the 
right. 

Use.  The  principal  use  of  evolution  in  arithmetic  is  in 
connection  with  a  few  problems  in  mensuration  involving 
right-angle  triangles,  areas,  and  volumes. 

180.  Logarition.  Logarition  is  the  process  of  finding 
the  number  of  times  the  equal  factor  occurs  from  the 
product  and  the  equal  factor.  The  product  is  the  antilog- 
arithm,  the  number  of  times  the  equal  factor  occurs  is 
the  logarithm,  and  the  equal  factor  is  the  base. 

ILL.  How  many  times  must  2  be  used  as  a  factor  to  make  8  ?  Or, 
to  what  power  must  2  be  raised  to  make  8?  This  question  is  written, 
what  is  log  82?  It  is  read,  what  is  the  logarithm  of  8  in  the  system 
whose  base  is  2?  Ans.  3. 


130  LESSON   25.    LOGARITION  §181 

Use.  Tables  have  been  prepared  stating  the  powers  to 
which  10  must  be  raised  to  produce  all  numbers.  By 
their  aid,  numbers  may  be  multiplied,  divided,  raised  to 
powers,  and  depressed  to  roots  with  marvelous  speed  and 
ease.  Logarithms  were  discovered  by  Napier  in  the 
seventeenth  century.  It  is  said,  "  The  miraculous  powers 
of  modern  calculation  are  due  to  three  inventions  :  the 
Hindu  notation,  decimal  fractions,  and  logarithms." 

181.  Exercises.  1.  Extract  the  square  root  of  2  true  to  two  deci- 
mal places  by  trial.  2.  True  to  4  decimal  places  by  the  formula 
method.  3.  Show  inductively  that  if  an  integer  is  pointed  off  into 
periods  of  two  figures  each,  the  square  root  of  the  first  period  gives 
the  first  figure  of  the  root,  the  square  root  of  the  first  two  periods 
gives  the  first  two  figures  of  the  root,  and  so  on.  4.  Show  deduc- 
tively that  in  extracting  the  cube  root,  a  decimal  must  be  pointed  off 
from  the  left.  5.  Show  deductively  that  there  are  as  many  decimal 
places  in  a  root  as  there  are  decimal  periods  in  the  power. 


LESSON  26.     ALGEBRA  IN  ARITHMETIC 

182.  Needs.     The  needs  arise  of  a  briefer  notation,  and 
of  a  simpler  method  of  solving  equations  than  is  afforded 
by  arithmetic.     How  to  satisfy  these  needs  as  fully  as  is 
desirable  in  elementary  schools  is  discussed  in  this  lesson. 
See  §  49. 

183.  Numbers  by   Letters.      The   first  letters   of  the 
alphabet,  a,  6,  c,  are  used  for  known  numbers,  and  the  last 
letters,  #,  y,  z,  for  unknown  numbers. 

ILL.  "  If  you  take  a  certain  number,  multiply  by  12,  add  10, 
divide  by  2,  and  subtract  5,  the  result  is  24 ;  find  the  number.  If 
the  result  is  24  after  5  is  subtracted  what  is  the  number  ?  29.  If  the 
result  is  29  after  a  number  is  divided  by  2  what  is  the  number?  58. 
And  so  on  ;  the  answer  is  4. 

•'  A  shorter  way  is  to  represent  the  required  number  by  x.  Think, 
x,  12x,  12x  +  10,  Qx  +  5,  6x,  Qx  =  24,  x  =  4." 

184.  Signs  +  and  — .     A  quality  may  be  considered  in 
two  opposite  phases,  as  heat  and  cold,  up  and  down,  east 
and  west.     One  of  these  phases  is  called  positive  and  rep- 
resented by  '  +  '  ;  the  other  is  called  negative  and  repre- 
sented by  '  — . ' 

ILL.  T.  "  Let  us  agree  that  opposites  may  be  represented  by  the 
signs,  '  +  '  and  '  —  ',  and  called  positive  and  negative. 

"  Name  two  opposites.  To  the  right  and  to  the  left ;  they  may  be 
represented  by  '  +  '  and  '  — '.  What  would  +  5  in.  mean?  5  in.  to 
the  right.  -  3  in.?  3  in.  to  the  left. 

"  Have  you  had  any  such  use  of  '  +  '  and  <  —  '  before  ?  Are  not 
addition  and  subtraction  opposites?  Is  not  one  represented  by  '  +' 
and  the  other  by  '  -  '  ?  " 

131 


132  LESSON  26.    ALGEBRA  IN  ARITHMETIC  §  185 

185.  Coefficients.     A  number  may  be  used  several  times 
as  an  addend  or  several  times  as  a  subtrahend.     The  num- 
ber of  times  is  called  the  coefficient.     A  positive  coefficient 
shows  how  many  times  a  base  is  used  as  an  addend  ;  a 
negative  coefficient  shows  how  many  times  a  base  is  used 
as  a  subtrahend. 

.  ILL.  T.  "  Write  by  the  use  of  signs  that  a  is  to  be  added  3  times. 
a  +  a  +  a  is  written  +  3  a  or  3  a ;  when  the  sign  is  '+  '  it  is  usually 
omitted ;  +  3  is  called  a  positive  coefficient ;  coefficient  means  working 
together  with.  Write  that  a  is  to  be  subtracted  3  times.  —  a  —  a  —  a 
is  written  —  3  a ;  —  3  is  called  a  negative  coefficient. 

"  What  does  +  3  a  mean  ?  What  does  —  3  a  mean  ?  What  does  a 
positive  coefficient  show?  A  negative  coefficient? 

"  Have  you  had  any  such  use  before  ?  Does  not  3x4  mean  4  +  4 
+  4  or  that  4  is  to  be  added  3  times  ?  +3  may  be  regarded  as  the 
coefficient  of  4." 

186.  Numbers  with  Direction.     The  need  arises  of  per- 
forming the  fundamental  operations  upon  numbers  with 
direction. 

Addition.  To  add  when  the  signs  are  alike  write  the 
sum  and  use  the  common  sign  ;  to  add  when  the  signs  are 
unlike  write  the  difference  and  use  the  sign  of  the  greater. 

ILL.     T.  "  We  want  a  rule  for  adding  numbers  with  signs. 

+  5     -  5     +5     -5 
j-jj     -_2     -2     +  2 

"  Suppose  that  '  +  '  means  to  the  right  and  <  —  '  to  the  left.  Find 
the  sums  of  the  numbers  on  the  board  in  this  way.  I  draw  a  line 
and  take  the  point  A  for  the  starting  point.  I  will  add  +  5  and  —  2. 
+  5  means  5  places  to  the  right,  —  one,  two,  three,  four,  five;  —  2 
means  2  places  to  the  left,  —  one,  two;  my  last  mark  is  3  places  to 
the  right  of  A  or  +  3 ;  the  sum  of  +  5  and  —  2  is  +  3. 

"  What  seems  to  be  the  rule  for  addition  when  the  signs  are  alike  ? 
When  the  signs  are  unlike  ? 

"  Let  us  see  why  this  rule  holds.     With  like  signs,  a  number  of 


§  186  LESSON  26.    ALGEBRA  IN  ARITHMETIC  133 

operations  is  to  be  united  with  a  number  of  the  same  operations  and 
the  sum  must  denote  the  same  operation.  With  unlike  signs,  a  num- 
ber of  additions  cancels  the  same  number  of  subtractions  and  the  sign 
of  the  result  is  the  sign  of  the  greater  coefficient." 

Subtraction.  To  subtract  change  the  sign  of  the  sub- 
trahend and  proceed  as  in  addition. 

It  would  be  very  confusing  if  a  minus  sign  were  written 
before  each  subtrahend  to  indicate  subtraction.  _;£*  would 
be  a  monstrosity.  See  §  74. 

ILL.   T.  "  We  want  a  rule  for  subtracting  numbers  with  signs. 

+  2         -2         -2         +2 
+_5        _-5         +_5         -5 

"  If  you  go  to  the  south,  what  do  you  do  to  your  position  ?  You 
subtract  from  your  north  position  and  add  to  your  south  position. 
If  you  go  to  the  north,  what  do  you  do  to  your  position  ?  You  sub- 
tract from  your  south  position  and  add  to  your  north  position.  It 
seems,  then,  that  to  subtract  a  number  with  either  sign  is  to  add  the 
number  with  its  sign  changed.  Apply  this  rule  to  the  numbers  on 
the  board.  To  subtract  -f  5  from  +  2,  change  the  subtrahend  and 
proceed  as  in  addition.  What  is  the  rule  for  subtraction? 

"  Let  us  look  at  this  in  another  way.  To  subtract  +  5  is  the 
opposite  of  to  subtract  —  5 ;  the  opposite  of  to  subtract  —  5  is  to  add 
—  5 ;  hence  to  subtract  +  5  is  to  add  —  5." 

Multiplication.  The  product  of  like  signs  is  plus  ;  the 
product  of  unlike  signs  is  minus. 

ILL.   T.  "  We  want  a  rule  for  multiplying  numbers  with  signs. 

"  —  3  x  —  4  means  that  —  4  is  to  be  subtracted  3  times.  —  4  from 
0  =  +  4  ;  -  4  from  +  4  =  +  8 ;  -  4  from  +  8=+ 12;  -3x-4 
=  +  12.  In  the  same  way,  find  the  value  of  +  3  x  -1-4.  Finish  the 
sentence  ;  the  product  of  numbers  with  like  signs  is . 

"  In  the  same  way,  find  the  value  of  +  3  x  — 4;  of  —  3  x  +4. 
Finish  the  sentence;  the  product  of  numbers  with  unlike  signs 
is ." 

Division.  The  quotient  of  like  signs  is  plus  ;  the  quo- 
tient of  unlike  signs  is  minus. 


134  LESSON  26.    ALGEBRA  IN  ARITHMETIC  §187 

ILL.    T.  "  We  want  a  rule  for  dividing  numbers  with  signs. 

"  Either  of  two  factors  is  equal  to  their  product  divided  by  the 
other  factor.  +  3  x  —  4  =  —  12.  From  this  law  what  must  be  —  12 
-..  _  4  ?  _  12  .j.  +  3  ? 

"  -  3  x  -  4  =  +  12.     What  must  be  +  12  -4-  -  3?     +  12  -4-  -  4? 

«  +  3  x  +  4  =  +  12.     What  must  be  +12  -*•  +  3  ?     +12  +  +  4  ? 

"  Finish  the  sentences ;  the  quotient  of  numbers  with  like  signs  is 
;  the  quotient  of  numbers  with  unlike  signs  is ." 

187.  Removing  Parentheses.     To  remove  a  parenthesis, 
multiply  every  term  within  it  by  its  coefficient. 

ILL.  T.  "  A  parenthesis  about  an  expression  means  that  every  term 
within  it  is  to  be  subjected  to  the  same  operation.  Thus,  —  5(6z  —  2) 
means  that  Qx  —  2  is  to  be  multiplied  by  —  5.  Remove  the  parenthe- 
sis; —  5(6x  —  2)  =  —  30x  +  10.  What  is  the  rule  for  removing  a 
parenthesis  ? 

"  A  bar  is  sometimes  used  in  place  of  a  parenthesis.  Find  the 
value  of  —  Qx  —  2.  This  means  that  6x  —  2  is  to  be  multiplied  by 
-  1.  -  Qx-  2  =  -  Qx  +  2." 

188.  From  nx  to  find  x.     Given  the  value  of  nx  to  find  z, 
divide  both  members  of  the  equation  by  the  coefficient  of  x. 

ILL.     T.   "-3z  =  -12.     How  shall  we  find  the  value  of  x  ? 

"  To  find  x  we  must  divide  —  3  x  by  —  3.  Dividing  both  members 
of  an  equation  by  the  same  number  cannot  affect  the  balance.  Divide 
both  members  of  the  equation  by  the  coefficient  of  a:,  x  =  4." 

189.  From  x"  to  find  x.     Given  the  value  of  xn  to  find 
a;,  extract  the  nili  root  of  both  members  of  the  equation. 

ILL.     T.   "x2  =  16.     How  shall  we  find  the  value  of  a;? 

"  To  find  x  we  must  extract  the  square  root  of  x2.  Extracting  the 
same  root  of  both  members  of  an  equation  cannot  affect  the  balance. 
Extract  the  square  root  of  both  members,  x  —  4.  x3  =  27 ;  find  x." 

190.  Transposing.     To  transpose  a  term  from  one  mem- 
ber of  an  equation  to  the  other,  change  its  sign. 

ILL.  T.  "2x  —  3  =  9.  It  is  necessary  to  transpose  —  3  to  the 
right-hand  member  of  the  equation  so  that  2  x  alone  will  be  in  the 
left-hand  member.  How  shall  we  do  it  ? 


§191  LESSON   26.    ALGEBRA   IN  ARITHMETIC  135 

"What  must  be  added  to  —  3  to  make  zero?  +  3.  Adding  the 
same  number  to  both  sides  cannot  affect  the  balance.  Adding  +  3  to 
both  sides  what  do  you  get?  2  x  —  9  +  3.  What  is  the  rule?" 

191.  Clearing  of  Fractions.  To  clear  of  fractions,  mul- 
tiply both  members  of  an  equation  by  the  least  common 
denominator. 

ILL.     T.   "  —  —  10  =  —  .     It  is  necessary  to  clear  of  fractions. 

"What  is  the  least  common  denominator?  Multiplying  both 
members  of  an  equation  by  the  same  number  cannot  affect  the  balance. 
Multiply  both  members  by  6.  Say,  2  is  contained  in  6,  3  times  ;  3  times 
3  x  is  9  x  ;  6  times  —  10  is  —  60  ;  3  is  contained  in  6,  2  times  ;  2  times 

2  x  is  4  x.     Ans.   9  x  —  60  =  4  x.     What  is  the  rule  for  clearing  of 
fractions  ? 

"Clear  of  fractions  (I)  on  the  board.     The  L.  C.  D.  is  5;  5  times 

3  x  is  15  x  ;  5  is  contained  in  5,  1  time  ;  1  times  —  1  is  —  1  ;   —  1  times 
6  a:  is  —  6  x  ;   —  1  times  —  2  is  +  2  ;  5  times  4  is  20.      .4ns.  15  x  —  6  x 

6  x  —  2 
+  2=  20.     I  want  you  to  see  why  we  say,  '  1  times  —  1.'     -- 

means  that  6  x  —  2  is  to  be  multiplied  by  —  1  (§  187)  ;  the  1  is  not 
written  but  must  be  understood. 

"  Clear  of  fractions  (II)  on  the  board.  Look  out  for  the  '  —  ' 
before  the  fraction.  5  is  contained  in  10,  2  times  ;  2  times  —  3  is  —  6  ; 
—  6  times  x  is  —  6  x  ;  —  6  times  -  4  is  -f  24." 


192.  Discussion.  Only  enough  of  algebra  should  be 
taught  in  the  elementary  schools  to  give  pupils  something 
of  an  idea  of  its  power  and  to  provide  for  the  indirect 
cases  in  percentage  and  interest  and  for  the  formulae  of 
mensuration.  The  latter,  for  \TT!)Z  and  ^TrD3,  require 
the  handling  of  equations  with  the  2d  power  of  x  and  no 
other  power  of  x,  and  equations  with  the  3d  power  of  x 
and  no  other  power  of  x. 


136  LESSON  26.    ALGEBRA  IN  ARITHMETIC  §  193 

193.  Exercises.  1.  By  counting  as  in  §  186,  find  the  sum  of  —  5 
and  +  2.  2.  If  income  is  constant,  savings  and  spendings  are  opposites ; 
what  is  the  effect  on  savings  of  subtracting  from  spendings  ?  3.  What 
is  the  effect  on  spendings  of  subtracting  from  earnings?  4.  In 
equation  (II)  of  §  191  find  the  value  of  x  and  explain  every  step. 
6.  Solve  the  equation,  \  wD2  =  78.54,  and  explain  every  step  (count 
•tr  as  3.1416).  6.  Solve  the  equation,  $*-Z>8=  113.0976,  and  explain 
every  step. 


LESSON   27.     PERCENTAGE 

194.  Development.     In  business  a  part   is   usually  ex- 
pressed as  a  number  of  the  hundred  equal  portions  of  a 
whole.      Hundredths  is  called  per  cent  and  the  fraction  is 
expressed  by  aid  of  the  symbol,  %. 

ILL.  T.  "  We  are  going  to  consider  another  way  of  expressing 
hundredths. 

"  Write  6  hundredths  as  a  common  fraction.  T^.  Write  small 
circles  in  place  of  the  numerator  and  denominator  and  write  6  to  the 
left.  6  %.  This  means  that  the  numerator  is  6  and  the  denominator 
100;  it  is  read  6  per  cent." 

195.  Reductions.      When  the  denominator  of  a  fraction 
is  2,  3,  4,  5,  6,  or  8,  it  is  customary  to  use  the  common 
fraction  in  place  of  its  per  cent  equivalent.     Pupils  should 
find  the  equivalents  and  master  them  so  thoroughly  that 
an  expression  in  either  form  will  suggest  the  other  in- 
stantly without  any  form  of  computation. 


3 

83  J% 

i       62*% 

* 

75  % 

f 

50% 

I       16|  % 

5 
I 

87*% 

i       12*% 

I 

33^  % 

3 

60% 

i      40  % 

4 

25  % 

1       66f% 

2 

37i% 

i 

20% 

fr      80  % 

ILL.  T.  "  I  asked  you  to  bring  in  on  paper  the  per  cent  equivalents 
of  the  business  fractions.  Here  they  are  on  this  chart.  As  I  point 
call  the  equivalent  instantly  (he  points  at  the  rate  of  one  a  second). 
You  fail  on  £,  f,  and  £.  Give  these  special  attention ;  you  must  not 
compute  their  values." 

196.  Terms.  In  the  full  expression  of  a  part  there  are 
three  terms,  the  whole,  the  fraction  of  the  whole,  and  the 

137 


138  LESSON  27.    PERCENTAGE  §  197 

part.  They  are  called  base,  rate,  and  percentage.  The 
base  plus  the  percentage  is  the  amount  ;  the  base  minus 
the  percentage  is  the  difference. 

ILL.  T.  "3  =  6  %  of  50 ;  53  =  50  +  6  %  of  50 ;  47  =  50  -  6  %  of  50. 
3  is  the  percentage ;  6  %,  the  rate ;  50,  the  base ;  53,  the  amount ;  47, 
the  difference." 

197.  Direct  Cases.  The  direct  cases  are  to  find  the  part, 
the  whole  plus  the  part,  and  the.  whole  minus  the  part, 
from  the  whole  and  the  fraction  of  the  whole  —  P,  A,  D 
from  B  and  R  (p.  7). 

At  least  99%  of  all  problems  in  percentage  outside  of 
the  schoolroom  are  of  these  types.  Pupils  should  work 
with  them  for  a  long  time  before  considering  the  other 
types,  and  until  they  can  find  results  rapidly  and  accu- 
rately both  without  the  pencil  and  with  the  pencil. 

It  is  best  both  for  mental  and  for  written  work  to  find 
1  %  and  to  modify  the  result.  It  is  worth  while  in  busi- 
ness problems  to  be  able  to  call  the  facts  in  the  following 
table  without  stopping  to  compute  them. 

1  %  of   f  100  =  $  1  1  %  of   $  10,000  =  $  100 

1  %  of  1 1000  =  $10  1  %  of  $  100,000  =  $  1000 

1  %  of  $  1,000,000  =  $  10,000 

Mental.  Before  taking  up  written  exercises  pupils 
should  gain  facility  in  mental  computations.  It  should 
be  remembered  that  the  pencil  is  required  only  when  the 
operations  are  difficult.  See  §  70. 

Exercises  like  the  following  are  as  valuable  in  prepara- 
tion for  the  solution  of  problems  as  examples  in  abstract 
multiplication  and  division. 

ILL.  Fractions.  T.  "What  is  f  of  30?  What  is  the  result  when 
30  is  increased  by  f  of  itself  ?  What  is  the  result  when  30  is  dimin- 
ished by  §  of  itself?  What  is  the  result  when  a  no.  is  increased 


§  197  LESSON  27.    PERCENTAGE  139 

by  §  of  itself  ?     -  no.     What  is  the  result  when  a  no.  is  diminished 
by  |  of  itself?     f  no." 

ILL.  Per  Cents.  T.  «  What  is  6  %  of  $  500  ?  Find  1  %  and  multiply 
by  6;  think,  1  %  of  $500  is  $5,  and  say,  830.  What  is  the  result 
when  $500  is  increased  by  6%  of  itself?  "\Vhat  is  the  result  when 
$500  is  diminished  by  6%  of  itself?  What  is  the  result  when  a  no. 
is  increased  by  6  %  of  itself?  106%  no.  What  is  the  result  when  a 
no.  is  diminished  by  6  %  of  itself  ?  94  %  no." 

ILL.  Equivalents.  T.  "  What  is  87£  %  of  $  48  ?  Think,  \  of  $  48  is 
$  6,  and  say,  $  42.  What  is  the  result  when  $  48  is  increased  by 
87^%  of  itself?  What  is  the  result  when  $48  is  diminished  by 
87£  %  of  itself  ?  What  is  the  result  when  a  no.  is  increased  by  87 £  %  of 
itself?  —  no.  What  is  the  result  when  a  no.  is  diminished  by 
87* %  of  itself?  I  no." 

ILL.  Frac.  Per  Cents.  T. ','  You  were  asked  to  memorize  the  facts  on 
the  board.  Clean  the  board.  What  is  1  %  of  $  10,000  ?  of  $  1,000,000  ? 
(he  drills  expecting  an  answer  each  second).  What  is  $  %  of  $  10,000  ? 
Think,  $  100,  and  say,  $  12.50."  (Such  exercises  are  valuable  because 
brokerage  is  usually  reckoned  as  |  %  of  the  par  value.) 

Written.  Examples  of  the  same  nature  as  the  foregoing 
should  be  given  to  afford  practice  when  the  computations 
cannot  be  readily  made  without  the  pencil.  Attention  is 
called  to  the  value  of  finding  1  %  as  the  first  step. 


$2.86X50 
37 


20055 
8595 
$  106.005 


$286.50 
.37 


20055 


$  106.005 


ILL.  To  find  37  %  of  $286.50,  we  move  the  decimal  point  to  find 
1  %,  and  then  multiply  by  37.  This  practice  conforms  to  the  plan  in 
mental  work  and  assists  in  fixing  the  place  of  the  decimal  point 


140 


LESSON  27.    PERCENTAGE 


198 


Since  30  x  2  is  60,  the  answer  cannot  be  so  small  as  $  10  nor  so  great 
as  $1060 ;  it  must  be  $  106.  As  a  check,  the  pupil  should  be  required 
to  fix  the  point  as  just  suggested,  and  also  by  counting  the  places. 

198.    Indirect  Cases.     The  indirect  cases  usually   pre- 
sented in  arithmetic  are  the  inverse  of  the  direct  cases. 


DlBEOT 

P  from  B  &  R ;  P  =  BR 

A  from  B  &  R;  A  =  B  +  BR 
D  from  B  &.  R;  D  =  B  -  BR 


INDIRECT 

R  from  B  &  P;  R  =  - 
B 


B  from  P  &  R ;  B  - 
B  from  A  &  R ;  B  = 
B  from  D  &  R ;  B- 


R 


1  +  R 

D 
1-  R 


ILL.  T.  "  You  may  copy  from  the  board  the  cases  which  you  have 
already  had.  They  are  called  direct.  From  the  first  case  write  a  new 
problem  in  which  the  answer  to  the  first  is  given  in  the  second  and 
the  required  term  in  the  second  is  6  % ;  do  the  same  making  the  re- 
quired term  50  (§  46).  From  the  2d  and  3d  direct  cases,  do  the 
same,  making  the  answer  the  given  term  and  50  the  required  term. 
These  four  new  cases  are  called  indirect." 


DIRECT 

What  is  6%  of  50?     .4ns.  3 

What  is  50  +  6%  of  50  ?    Am.  53 
What  is  50  -  6  %  of  50  ?    A  ns.  47 


INDIRECT 


f  3  is  what  %  of  50? 

J3  is  6%  of  what? 

53  is  6  %  more  than  what? 

47  is  6  %  less  than  what  ? 


Solutions  in  General.  The  indirect  cases  can  be  solved 
by  algebra,  by  formula,  by  rule,  or  by  analysis.  The  2d 
and  3d  cases  can  also  be  solved  by  proportion,  or  by 
variation  (Lesson  11). 


§198 


LESSON   27.    PERCENTAGE 


141 


By  algebra,  the  values  of  the  known  terms  may  be  sub- 
stituted in  the  formula  for  the  direct  relation  and  the 
equation  solved,  or  the  required  term  may  be  represented 
by  x  and  the  equation  formed  by  reasoning.  See  §  49. 

By  formula,  the  required  term  is  made  the  left-hand 
member  of  an  equation  derived  from  the  direct  formula 
before  the  substitutions  are  made.  See  §  50. 

By  rule,  the  translation  of  the  formula  obtained  by  the 
last  method  is  stated  from  memory.  See  §  51. 

By  analysis,  the  problem  is  separated  into  its  simple 
problems.  See  §  36. 

Each  method  has  its  advocates.  The  author  prefers  the 
analysis  method,  or  the  2d  algebra  method,  because  by 
them  it  is  not  necessary  to  decide  upon  the  technical  name 
for  each  term,  a  decision  which  is  sometimes  quite  per- 
plexing. Thus,  in  the  problem  below,  by  these  methods 
it  is  not  necessary  to  decide  that  47  is  the  difference. 

ILL.  If  a  number  is  diminished  by  6  %  of  itself,  the  result  is  47. 
What  is  the  number? 


1ST 

ALGEBRA                                                         2n 

ALGEBRA 

A  47 

D  =  B-BR           No.  dec.,  47 

Let  x  =  no. 

R,  6% 

47  =  £-.06  B         Dec.,  6%  no. 

.06  x  =  dec. 

Z?      Q 

X>)  f 

47  =  .94  B                  JJQ   v 

.94  x  =  no.  dec. 

' 

B      47      50 

.94x  =  47 

5  =  50.  .4ns. 

•9*                    No.,  50.   ^ns 

z  =  50 

FORMULA 

RULE 

D,  47 

D=B-BR 

A  47 

Given  the  differ- 
ence and  the  rate 

tf,  6% 

J-J  ^=  Ij  (1  —  Xt  ) 

ft     Q  01 

to  find  the  base, 

5'?_ 

j^  Jy 

B,7 

divide  the  differ- 
ence by  1  minus 

l  —  R 

5,  50.   ;lrw. 

=  ^  =  50 

B,  50.  Ans. 

the  rate. 

47  +  .94  =  50 

142  LESSON  27.    PERCENTAGE  §  198 

Analysis  in  General.  It  is  convenient  to  designate  the 
three  direct  cases  as  Case  1,  and  the  four  indirect  cases  as 
Case  2,  Case  3,  Case  4,  and  Case  5. 

ILL.    Case  2.     T.  "  12  is  what  part  of  20  ?    Ans.  ^  of  20  or  .f  of  20. 

"  12  is  what  %  of  20  ?    Ans.  12  is  ^  of  20  or  |  of  20  or  60  %  of  20." 

ILL.  Case  3.  1.  "12  is  f  of  what?  Ans.  If  12  is  f  110.,  what  is  £ 
no.  ?  4.  If  4  is  I  no.,  what  is  f  no.  ?  #0. 

"12  is  3%  of  what?  Ans.  If  12  is  3%  no.,  what  is  1%  no.?  4. 
If  4  is  1  %  no.,  what  is  100  %  no.  ?  ^00." 

ILL.  Cases  J+fy  5.  T.  "  What  no.  increased  by  |  of  itself  becomes 
14?  Ans.  If  a  number  is  increased  by  f  of  itself,  what  does  it  be- 
come ?  I  no.  If  |  no.  is  14,  etc. 

"What  no.  increased  by  6  %  of  itself  becomes  53?  Ans.  If  a  no. 
is  increased  by  6  %  of  itself,  what  does  it  become  ?  106  %  no.  If  106  % 
no.  is  53,  what  is  1  %  no.  ?  £.  If  1  %  no.,  is  £,  etc. 

"  What  no.  decreased  by  f  of  itself  becomes  21  ? 

"  What  no.  decreased  by  6  %  of  itself  becomes  47  ?  " 

Analysis  in  Particular.  Cases  4  and  5  may  be  regarded 
as  another  way  of  stating  Case  3  and  may  be  treated  as 
Case  3.  Pupils  should  practice  changing  from  one  form 
of  expression  to  the  other. 

ILL.  T.  "  Change  to  an  expression  without  more  or  less :  16  is  J 
more  than  12.  Ans.  16  is  f  of  12.  8  is  |  less  than  12.  Ans.  8  is  f 
of  12.  A's  money  is  £  more  than  B's.  A  ns.  A's  money  is  f  as  much 
as  B's.  In  a  mixture  of  water  and  vinegar  the  water  is  \  more  than 
the  vinegar.  Ans.  In  a  mixture  of  water  and  vinegar  the  water  is  f 
as  much  as  the  vinegar. 

"  Change  to  an  expression  with  more  or  less :  A's  weight  is  f  of  B's. 
Ans.  A's  weight  is  f  less  than  B's.  A's  salary  this  year  is  f  as  much 
as  last  year.  Ans.  A's  salary  this  year  is  \  more  than  last  year. 

"  Solve  as  Case  3.  If  a  no.  is  increased  by  6  %  of  itself  the  result  is 
53.  What  is  the  no.  ?  Ans.  If  106  %  of  a  no.  is  53,  what  is  1  %  no.  ? 
etc.  In  a  mixture  of  14  gal.  of  water  and  vinegar  the  water  is  \  more 
than  the  vinegar.  How  much  is  the  vinegar?  Ans.  If  the  water  is 
f  as  much  as  the  vinegar,  how  much  is  the  mixture  ?  |  as  much  as 
the  vinegar.  If  |  as  much  as  the  vinegar  is  14  gal.,  etc." 


§199 


LESSON   27.    PERCENTAGE 


143 


Analysis  Written  Problems.  A  problem  can  be  stated 
in  full  from  the  proper  expression  of  what  is  given  and 
what  is  required. 


CASE  2.    Had  210  sheep  and  sold 
54.     Sold  what  %  flock  ? 

F,  210  sh 

Sold,  84  sh 

Sold,  ?%F  Jft  =  H 


Sold,  40  %F  ,4ns. 


=  40% 


CASE  3.     Sold  84  sheep  which  was 
40  %  flock.    Flock  what  ? 

Sold,  84  sh  or  |  F 
F,?  42 
210 


1  F,  42  sh 

F,  210  sh  Ans. 


CASE  4.    Bought  40%  flock ;  theu 
had  294  sheep.    Flock  what  ? 

Aft.  P,  294  sh  or  £  F 
F,? 


42 
210 


t  F,  42  sh 

F,  210  sh  .4ns. 


CASE  5.      Sold  40%   flock ;    then 
had  126  sheep.    Flock  what  ? 

Aft.  S,  126  sh  or  f  F 
F,?  42 

210 

t  F,  42  sh 
F,  210  sh  ^4ns. 


Case  2.  If  he  had  210  sheep  and  sold  84  sheep,  what  per  cent  of 
his  flock  did  he  sell  ?  ^  or  ^  or  |  or  40%. 

Case  4.  If  he  purchased  §  as  many  sheep  as  were  in  the  flock,  how 
many  sheep  did  he  then  have  ?  ?  flock.  If  \  flock  is  294  sheep,  etc. 

199.  Exercises.  1.  Try  the  reductions  of  §  195.  Can  you  call 
one  a  second?  2.  Can  you  name  the  results  in  §  197  in  5  seconds? 
3.  In  §  198  show  how  each  of  the  indirect  formulas  is  derived  from 
its  direct  form.  4.  State  the  problem  of  Case  3,  p.  143.  6.  Solve  it 
by  the  1st  algebra  method.  6.  Solve  it  by  the  2d  algebra  method. 
7.  Solve  it  by  the  formula  method.  8.  Solve  it  by  rule.  9.  State 
with  reasons  which  method  you  prefer.  10.  Which  of  the  five  cases 
should  receive  chief  attention  ?  Why  ?  11.  How  many  cases  are 
there  in  percentage  ?  See  p.  7.  12.  State  a  problem  for  every  case. 


LESSON   28.     PERCENTAGE 

200.  Important  Suggestion.     In  solutions,  what  the  per 
cent  is  of  should  be  expressed  after  every  per  cent.     It  is 
usually  omitted  in  the  statement  of  a  problem,  and  if  it  is 
omitted  in  the  solution  also,  clear  thinking  is  impossible. 

201.  Profit  and  Loss.     Unless  otherwise  specified  gain 
or  loss  is  reckoned  as  some  per  cent  of  the  cost. 

ILL.  Development.  T.  "  Shall  we  regard  gain  and  loss  as  some 
per  cent  of  the  cost  or  some  per  cent  of  the  selling  price  ?  As  some 
per  cent  of  the  cost  because  the  cost  comes  first ;  a  man  cannot  sell  a 
thing  until  he  gets  it.  A  merchant  sells  goods  at  a  gain  of  40%. 
Supply  the  omission.  Ans.  A  merchant  sells  goods  at  a  gain  of  40% 
of  the  cost." 

ILL.  Direct  Cases.  T.  "  Goods  that  cost  50  <f>  are  sold  at  a  gain  of 
40%.  What  is  the  gain?  the  selling  price?  Ans.  If  the  cost  is  50^ 
and  the  gain  is  40  %  of  the  cost,  what  is  the  gain  ?  20  p.  Be  sure  to 
use  after  40  %  what  40  %  is  of. 

"  Butter  that  costs  24  f>  is  sold  at  a  loss  of  12£  %.  What  is  the  loss  ? 
the  selling  price  ?  " 

ILL.  Indirect  Cases.  T.  "  Make  the  statement  without  the  use  of 
the  words,  gain  or  loss.  An  article  is  sold  for  60^  at  a  gain  of  33|%. 
Ans.  An  article  is  sold  for  60^  or  for  $  cost.  An  article  is  sold  for 
40^  at  a  loss  of  16f%.  Ans.  An  article  is  sold  for  40^  or  for  f  cost. 

"  A  book  that  cost  $  6  was  sold  for  $  4.     What  per  cent  was  lost  ? 

"  On  goods  sold  at  a  profit  of  66f  %  the  gain  was  $24.  What  was 
the  selling  price  ?  A  ns.  If  f  cost  was  $  24,  etc. 

"  Goods  are  sold  for  $2.50  at  a  gain  of  25%.     What  is  the  cost? 

"  A  coat  was  sold  for  $  15  at  a  loss  of  16|%.  What  was  the  loss  in 
dollars  ?  " 

Written  Problems.  Be  sure  to  write  after  each  per  cent 
expression  what  it  is  of. 

144 


§201 


LESSON  28.    PERCENTAGE 


145 


CASE  1.  A  man  bought  goods  for 
$  528.60  and  sold  them  at  a  gain  of 
6  %.  What  was  the  gain  ? 

C,  $  528.60 
G,  6%C 
G,? 


5.28x60 
6 


G,  $  31.72  Ans. 


31.716 


CASE  2.  A  man  bought  goods  for 
$528.(>0  and  sold  them  at  a  gain  of 
$31.716.  What  %  did  he  gain? 

C,  $528.60 

G,  $31.716  .06 

G,  ?  %  C  528x6.)31*7.16 
31  716 

G,  6%  C   Ans. 


CASE  3.  A  man  sold  goods  at  a 
gain  of  6%  or  at  a  gain  of  -$31.716. 
What  was  the  cost  ? 

G,  $31.716  or6%  C 
C,? 

6)31.716 
C,  $528.60  Ans. 


CASE  4.  A  man  sold  goods  for 
$560.316  at  a  gain  of  6%.  What 
was  the  cost  ? 

S,  $560.316  or  106%  C 

C>  '•  5.286 

106)560.316 
C,  $528.60  Ans.  530 


EXPL.  Case4-  If  106%  cost  is  $560.316,  what  is  1%  cost?  $5.286. 
If  1  %  cost  is  $  5.286,  what  is  the  cost  ?  $  528.60. 

Difficult  Problems.  Problems  involving  two  or  more 
bases  are  often  given  in  examinations  of  teachers.  They 
should  not  be  taught  in  the  elementary  schools.  The 
secret  of  success  in  their  solution  is  to  state  what  the  per 
cent  is  of  after  each  per  cent. 

ILL.  1.  At  what  per  cent  must  a  merchant  mark  goods  so  that 
he  can  make  a  discount  of  10  %  and  yet  make  a  gain  of  26%? 

As  Case  4  and  Case  5  this  means,  At  what  per  cent  of  the  cost 
must  a  merchant  mark  goods  so  that  he  can  make  a  discount  of  10% 
of  the  marked  price  and  yet  make  a  gain  of  26  %  of  the  cost  ? 

As  Case  3  this  means,  At  what  per  cent  of  the  cost  must  a  mer- 
chant mark  goods  so  that  he  can  sell  them  for  90%  of  the  marked 
price  or  for  126  %  of  the  cost  ? 

If  90%  marked  price  is  126%  cost,  what  is  1%  marked  price?  etc. 

ILL.  2.  A  man  sold  two  articles  at  the  same  price.  On  one  he 
gained  20%  and  on  the  other  he  lost  20%.  What  per  cent  did  he 
gain  or  lose  on  the  whole? 


146  LESSON  28.    PERCENTAGE  §202 

This  means,  On  the  first  he  gained  20  %  cost  of  first  and  on  the 
second  he  lost  20%  cost  of  second.  What  per  cent  of  the  cost  of  both 
did  he  gain  or  lose  ?  As  Case  3  it  means,  He  sold  the  first  for  f  cost 
of  first  and  the  second  for  f  cost  of  second,  etc. 

The  cost  of  each,  the  cost  of  both,  the  selling  price  of  each,  and 
the  selling  price  of  both  must  be  expressed  by  the  same  unit. 
If  SP  is  f  Cp  what  is  C^  f  SP.  If  SP  is  $  C2,  what  is  C2?  £  SP. 
What  is  cost  of  both  ?  f  f  SP.  What  is  selling  price  of  both  ?  f  f  SP. 
What  is  the  loss  on  both?  ^  SP.,  etc. 

202.  Commission.  When  an  agent  buys,  his  commission 
is  some  per  cent  of  his  buying  price  ;  when  he  sells,  his 
commission  is  some  per  cent  of  his  selling  price  ;  when  he 
collects,  his  commission  is  some  per  cent  of  his  collection. 

ILL.  T.  "  A  person  may  engage  an  agent  to  transact  business.  In 
what  capacities  may  the  agent  serve  ?  He  may  buy,  or  sell,  or  collect. 

"  Give  me  an  illustration  of  an  agent  buying.  A  grain  dealer  in 
N.  Y.  City  employs  an  agent  in  Chicago  to  purchase  1000  bu.  of  wheat. 
The  agent's  commission  is  1  %.  What  is  the  commission  1  %  of  ?  1  % 
of  what  the  agent  pays  for  the  wheat.  Why  ? 

"  Give  me  an  illustration  of  an  agent  selling.  A  farmer  sends  100 
bbl.  of  apples  to  a  merchant  in  N.  Y.  City,  directing  him  to  sell  them. 
The  agent's  commission  is  10%.  10%  of  what?  Of  what  the  mer- 
chant sells  them  for.  Why? 

"  Give  me  an  illustration  of  an  agent  collecting.  An  owner  of  an 
apartment  house  employs  an  agent  to  collect  the  rents.  The  agent's 
commission  is  2|  %.  2|  %  of  what  ?  Of  what  he  collects.  Why  ?  " 

Difficult  Problems.  Problems  in  which  the  commissions 
are  of  different  bases  are  often  given  in  examinations  of 
teachers.  They  should  not  be  taught  in  the  elementary 
schools.  The  secret  of  success  in  their  solution  is  to  state 
what  the  commission  is  of  after  every  per  cent. 

ILL.  An  agent  sells  goods  for  $4800,  charging  3£%  commission. 
After  paying  $25  charges  he  invests  the  balance  in  raw  material,  re- 
taining a  commission  of  2£  %.  How  much  does  the  agent  pay  for  raw 
material  ? 


§203  LESSON  28.    PERCENTAGE  147 

Agt'sSP,  $4800                          48.00*  102^)4619 

1st  com,  3J%  Agt's  SP  _3£ 

Ch,  $25                                  12  45.0634 

2d  com,  2£  %  Agt's  C                  144  205)9238. 

Agt's  C,  ?                                     156  820 

4800  1038 

1st  com,  $  156                           4644  1025 

Rem,  $4644  etc. 

Af.  ch,  $ 4619  or  102^  %  Agt's  C 

Agt's  C,  $ 4506.34  Ans. 

EXPL.  If  the  agt's  SP  is  $4800  and  the  1st  com.  is  3J%  of  agt's 
SP,  what  is  the  1st  com.?  If  the  agt's  SP  is  $4800  and  the  com. 
$156,  what  is  the  rern.?  If  the  rem.  is  $4644  and  the  charges  $25, 
what  is  left  after  the  charges  V  If  the  2nd  com.  is  2\  %  agt's  C,  what 
is  the  amount  with  the  com.?  If  102£  %  agt's  C  is  $4619,  etc. 

203.  Commercial  Discount.  The  net  price  is  the  con- 
tinued product  of  the  list  price  and  the  remainders  found 
by  subtracting  each  discount  from  100%. 

ILL.  Development.  T.  "  Manufacturers  and  wholesale  dealers  usu- 
ally publish  a  catalogue  of  their  goods  with  a  fixed  price  after  each 
article.  Here  is  such  a  list  (he  passes  it  around).  For  the  sake  of 
making  quick  sales  or  to  meet  the  market,  how  can  they  change  these 
prices?  It  would  be  very  expensive  to  publish  a  new  catalogue. 
They  offer  a  single  discount  or  several  successive  discounts  from  the 
list  price. 

"  The  first  discount  is  some  per  cent  of  the  list  price,  the  second  is 
some  per  cent  of  the  1st  remainder,  the  third  is  some  per  cent  of  the 
2d  remainder,  and  so  on. 

"  What  is  the  net  price  with  a  single  discount  of  25%?  Ans.  75% 
L ;  it  is  the  list  minus  25  %  list. 

"  What  is  the  net  price  with  two  discounts  of  25  %  and  20  %  ?  A  ns. 
60%  L.  The  rem.  after  the  1st  dis.  is  75%  L;  the  rem.  after  the  2d 
dis.  is  '  80  %  or  $  of  75  %  L '  or  60  %  L. 

"  What  is  the  net  price  with  three  discounts  of  25%,  20%  and  10%? 
,4ns.  54%  L.  The  rem.  after  the  2d  dis.  is  60%  L;  the  rem.  after 
the  3d  dis.  is  '  90%  or  T95  of  60%  L,'  or  54  %  L." 


148  LESSON  28.    PERCENTAGE  §204 

ILL.  Direct  Cases.  T.  "  What  is  the  net  price  of  a  bill  of  goods 
listed  at  $100  with  two  discounts  of  10%  and  5%?  Ans.  The  net 
price  is  95  %  of  90  %  L  or  85.5  %  L ;  85.5  %  of  $  100  is  $  85.50. 

"Work  it  in  another  way.  Ans.  The  1st  dis.  is  10%  of  $100  or 
$  10 ;  the  1st  rem.  is  $  100-  $  10  or  $  90 ;  the  2nd  dis.  is  5  %  of  $  90  or 
$4.50;  the  2d  rem.  is  $90 -$4.50  or  $85.50. 

"  What  is  the  difference  in  the  net  price  between  three  discounts  of 
20  %,  10  %,  and  5  %,  and  three  discounts  of  5  %,  10  %,  and  20  %  ?  A  ns. 
One  is  80  %  of  90  %  of  95  %  L,  and  the  other  is  95  %  of  90  %  of  80  %  L. 
The  product  is  the  same  in  whatever  order  the  multiplication  is 
performed.  There  is  no  difference." 

204.  Stocks.  Development.  Several  persons  may  form 
a  company  to  do  business  as  a  single  individual,  and  may 
become  incorporated  by  obtaining  a  legal  charter  from 
the  secretary  of  state.  Such  a  corporation  is  a  stock 
company,  their  holdings  is  stock,  the  equal  parts  into 
which  the  stock  is  divided  are  shares,  and  the  paper  show- 
ing how  many  shares  have  been  sold  at  'one  time  is  a 
certificate  of  stock. 

ILL.  July  1,  1911,  several  individuals  living  in  N".  Y.  City  obtained 
a  charter  under  the  name  of  the  Hygiea  Ice  Co.  for  the  production  of 
artificial  ice.  After  electing  John  Smith  president  and  Henry  Brown 
treasurer,  they  printed  2000  certificates  of  stock  and  bound  them  into 
books. 


CKR.  No. 

CER.  No. 

NEW  YORK  CITY, 

N.Y. 

No.  STT. 

-,  19 

No.  STT. 

c/rlLQs  i&  to  &L 

*i///V  Mat. 

DATE 

V     / 

isjs     vfl&     OU^Ti&l/ 

o-i                               aAa/ 

^ea/   ot 

NAME 

/ 

'uq^&d*  c?&&  (tLo-.f  of/LiA 

1J    0  !t/u         0si/h<tn? 

s&  /oy 

ADDRESS 

f  000   &'rlCl>1,&Q'  '     / 

0^i/i/  w^Lu/E'  ol-  /  a-^/., 

$/00. 

PRES. 

TBEAS. 

§204  LESSON  28.    PERCENTAGE  149 

They  sold  all  of  the  stock  at  $  80  a  share,  and  thus  obtained  a  cap- 
ital of  $80,000  for  the  development  of  the  business. 

A  certificate  duly  made  out  and  signed  is  given  to  each  pur- 
chaser, and  the  stub  duly  filled  in  is  retained  by  the  company. 

ILL.  Oct.  15,  191i,  the  company  sold  50  shares  to  George  W. 
Williams,  Yonkers,  N.  Y.,  and  issued  to  him  certificate  No.  100. 
Write  the  certificate  and  the  stub  and  detach  the  stub. 

If  the  company  is  prosperous,  it  issues  dividends  at 
equal  intervals,  as  annually  or  semiannually,  declared  as 
some  per  cent  of  the  par  value,  that  is,  some  per  cent  of 
the  value  printed  on  the  certificate. 

ILL.  During  the  first  year  the  company  made  $  10,000  clear  of  all 
expenses.  They  retained  $2000,  and  paid  out  $8000  in  dividends. 

What  per  cent  of  the  par  value  was  the  dividend?  What  was  the 
dividend  in  dollars  on  1  share?  How  much  was  Williams'  dividend? 
What  effect  on  the  market  value  of  the  stock  should  be  expected  from 
this  large  dividend  ? 

If  the  owner  of  a  certificate  of  stock  desires  to  sell,  he 
may  offer  it  to  different  individuals,  but  usually  he  takes 
it  to  a  dealer  in  stocks  (broker)  and  pays  a  commission 
(brokerage)  for  making  the  sale. 

ILL.   Oct.  15,  1912,  Williams  decided  to  sell  his  stock.     He  wrote 

on  the  back  of  his  certificate,  "  Transfer  to  the  order  of 

George  W.  Williams,"  and  gave  it  to  a  broker,  who  sold  it  to  Henry 
Wilson  for  $102  a  share.  The  broker  retained  $£  per  share  for 
brokerage  and  2  ^  a  share  for  state  tax,  and  sent  Williams  the  bal- 
ance. He  wrote  the  name  of  Henry  Wilson  on  the  line  left  for  the 
name  of  the  purchaser,  and  sent  the  stock  to  the  home  office,  where 
a  new  certificate  was  issued  and  sent  to  Henry  Wilson. 

Make  the  transfer  indorsement  on  the  certificate.  How  much  did 
Williams  make  on  his  investment  ?  Consider  cost,  dividend,  selling 
price,  brokerage,  and  state  tax. 

Preferred  and  Common.  Sometimes  a  company  issues 
stock  of  two  kinds,  preferred  and  common.  The  pre- 


150  LESSON  28.    PERCENTAGE  §204 

ferred  states  that  a  specified  dividend  will  be  paid  at  fixed 
intervals  if  the  earnings  of  the  company  warrant ;  the 
common  states  that  no  dividends  will  be  paid  until  all 
dividends  of  the  preferred  have  been  satisfied.  The  pre- 
ferred is  the  more  secure,  but  the  common  often  pays  the 
larger  dividends. 

Forms  of  Expression.  The  business  terms  used  in 
stocks  are  much  abbreviated,  and  are  always  some  per 
cent  of  the  par  value.  For  clear  thinking,  pupils  should 
be  required  to  use  the  unabbreviated  forms,  and  to  express 
each  term  as  dollars  a  share. 

BUSINESS  FORM  UNABBREVIATED  FORM 

A  6  %  dividend.  A  dividend  of  $6  a  share. 

5  %  stock.  Stock  that  pays  an  annual  divi- 

dend of  <$  5  a  share. 

Stock  @  90.  Stock  @  $  90  a  share. 

Stock  @  90  %.  Stock  @  $  90  a  share. 

Brokerage  |.  Brokerage  $  \  a  share. 

Brokerage  \  %.  Brokerage  $  \  a  share. 

Bought  at  a  premium  of  20  %.  Bought  @  $  120  a  share. 

Sold  at  a  discount  of  20  %.  Sold  @  $  80  a  share. 

$  6000  stock.  60  shares  of  stock. 

6  %  stock  yields  an  income  of  5  %.  Stock  pays  a  dividend  of  $  6  a  share 

or  5%  of  the  cost  of  a  share. 

Written  Problems.  In  the  statement  of  what  is  given 
and  what  is  required  it  is  best  to  use  the  unabbreviated 
forms. 

III.  4.  If  the  first  cost  of  1  sh.  is  $89f  and  the  br.  is  $$,  what  is 
the  entire  cost  of  1  sh.?  1 90.  If  the  inc.  on  1  sh.  is  $6,  on  how 
many  shares  is  the  inc.  $  540  ?  90.  If  the  cost  of  1  sh.  is  $  90,  what 
is  the  cost  of  90  sh.  ?  $  8100. 

III.  6.  If  5%  cost  is  $8,  what  is  1%  cost?  $1.60.  If  1%  cost  is 
$1.60,  what  is  the  cost?  $  160. 


205 


LESSON  28.    PERCENTAGE 


151 


ILL.  1.     What  is  the  cost  of  $4800 
stock  at  110,  brokerage  i? 

Mvl  sh,  $110 
Brlsh,$J  110, 

No.  sh,  48  48 

C,  ?  ~fi~ 

880 


C  1  sh  «110i 
C,  $5286 


•  ILL.  2.    How  many  sh.  at  20  %  dis. 
can  be  purchased  for  $5128,  br.  S  ? 

Mvlsh,  $80 
Brlsh,  9| 
C,  $5128 
No.  sh,  ? 


C    1  sh,  $80£ 
Sh,  64  Ans. 


80^)5128 

64 

641)41024 
3846 
2564 
2564 


ILL.  3.  What  income  will  be  ob- 
tained from  $1600  invested  in  5% 
stock  @  79J,  br.  J  %  ? 

Inv,  $1600 

Mv  1  sh,  $79$ 

Br  1  sh,  9| 

D  1  sh,  $5 

Td,  ?_ 

No.  sh,  20 
Td,  $100  Ans. 


ILL.  4.  How  much  must  be  inv. 
in  6  %  stock  @  89|,  br.  |,  to  get  an 
annual  inc.  of  $540? 

C  1  sh, 
D  1  sh,  $ 
Td,  $540 
Inv,  ? 


No.  sh,  90 

Inv,  $8100  Ans. 


ILL.  5.    What  percent  income  will 
I  receive  if  I  buy  6  %  stock  @  50? 

C  1  sh,  $50 
D  1  sh,  $6 
I),  ?  %  C 

10   v 

D,  12  %C  Ans. 


ILL.  6.    How  much  must  I  pay  for 
8  %  stock  to  make  5  %  ? 

Dl  sh,  $8  or  5%C 
C  1  sh,  ? 


C  1  sh,  $160  Ans. 


205.  Exercises.  1.  Explain  the  problems  above.  2.  Explain 
how  each  full  form  on  p.  150  is  derived  from  the  business  form. 
3.  Translate  the  statement  of  each  problem  on  p.  151  from  the  busi- 
ness form  to  the  full  form.  Thus,  ILL.  1.  What  is  the  cost  of  48 
shares  of  stock  at  $110  a  share,  brokerage  $  J  a  share?  4.  Consider 
the  par  value  $50  a  share  and  change  every  business  form  on  p.  150  to 
its  full  form.  Thus,  "  A  6 %  dividend  "  is  «  A  dividend  of  '$ 3  a  share." 


LESSON   29.     INTEREST 

206.  Negotiable  Notes.  Without  Interest.  In  order  to 
facilitate  the  transaction  of  business  one  person  frequently 
gives  to  another  a  written  promise  to  pay  a  given  sum  of 
money  at  a  given  time  and  place.  Pupils  should  write 
and  memorize  the  form  of  such  an  agreement ;  they  should 
be  careful  not  to  sign  their  names  to  business  papers  pre- 
pared in  school. 

ILL.  T.  "  July  1,  1912,  Henry  Jones  buys  a  bill  of  goods  of  John 
Smith  for  $1100.  Instead  of  paying  cash  he  gives  a  written  promise 
called  a  note,  agreeing  to  pay  the  bill  3  mos.  after  date  at  the  Tenth 
National  Bank.  In  order  that  this  note  may  be  negotiable  Jones 
makes  it  payable  to  the  order  of  John  Smith. 


$//00.00  N.  Y.  CITY,  N.  Y.,  fa      /,  19 

c5%-t*5£'  wu>n 
i&&  to     a,     to- 


at 

B&K&. 

ofo.  I 


"Copy  this  note  on  a  piece  of  paper  7"  x  3",  and  mark  it  No.  1 ; 
memorize  the  form.  Explain  the  features  of  the  note.  The  note 
must  state  the  date  and  place  where  it  is  drawn ;  the  date  and  place 
where  it  is  to  be  paid ;  that  value  has  been  received ;  and  by  whom, 
to  whom,  and  how  much  is  to  be  paid. 

"What  are  the  technical  terms?  Henry  Jones  is  the  drawer  or 
maker  or  payer ;  John  Smith  is  the  drawee  or  payee  ;  the  date  of  pay- 

152 


§  206  LESSON   29.     INTEREST  153 

ment  is  the  date  of  maturity ;  the  sum  for  which  the  note  is  drawn 
is  the  face;  the  amount  to  be  paid  at  the  date  of  maturity  is  the 
amount  at  maturity. 

"  What  is  the  date  of  maturity  ?  Oct.  1.  If  the  note  had  read  '  90 
days  after  date,'  what  would  have  been  the  date  of  maturity  ?  July 
91  or  Aug.  60  or  Sept.  29  (§  164).  What  is  the  amount  at  maturity? 
The  face  or  $1100." 

With  Interest.  If  a  note  without  interest  is  not  paid 
when  it  becomes  due,  interest  begins  with  the  date  of 
maturity.  Sometimes,  however,  the  maker  agrees  to  pay 
interest  from  the  date  of  the  note.  In  this  case,  with  in- 
terest or  with  interest  at  a  specified  rate  is  added  to  the 
note. 

ILL.  T.  "  By  the  sale  of  these  goods  Jones  expected  to  make  40% 
of  their  cost  or  $440.  He  was  willing,  therefore,  to  pay  something, 
interest,  for  the  use  of  the  money.  Suppose  he  wrote  the  note  for 
3  mo.  with  interest  at  4%.  Write  the  note  and  mark  it  No.  2. 


$/fOO.OO  N.  Y.  CITY,  N.  Y.,  fufy  /,  19  12 

a-it&i, 


to-      a-      to 


at  the- 


Bank,  with  i/nt&ieat  at  ty-  %. 
cA"o.  2  /"f&nvy 


"What  is  the  date  of  maturity?  Oct.  1.  What  is  the  period  for 
which  $1100  is  to  bear  interest?  Your  answer,  3  mos.,  is  contrary  to 
business  usage.  The  interest  period  is  the  actual  no.  of  days  from 
the  date  of  the  note,  July  1,  to  its  maturity,  Oct.  1,  or  92  days.  See 
§164. 

"If  'with  interest'  had  been  written  instead  of  'with  interest  at 
4% '  what  would  have  been  the  rate?  The  legal  rate  of  the  state  of 
New  York,  or  6%." 


154  LESSON   29.     INTEREST  §207 

207.  Interest  Computed.  The  cancellation  method  is  a 
favorite  with  teachers  because  it  is  easily  taught.  It 
should  be  made  the  basis  because  it  leads  at  once  to  valu- 
able modifications. 

Cancellation  Method.  No  rule  is  necessary.  The  pupil 
finds  the  simple  problems  and  their  answers.  See  §  39. 

ILL.  T.  "Look  at  Note  No.  2.  We  must  find  the  interest  of 
$1100  for  92  days  at  4%.  Before  solving  this  problem  we  will  solve  a 
few  others. 

"Find  the  interest  of  $204  for  1  yr.  5  mo.  17  da.  at  7%.  There 
are  many  ways  of  proceeding.  We  will  find  the  interest  for  1  da.  and 
then  multiply  by  the  no.  of  days.  How  many  days  shall  we  count  as 
a  year?  In  a  year  there  are  674  sec.  less  than  365 \  da.  (p.  112),  but  un- 
less otherwise  specified  360  da.  are  counted  to  the  year  in  computing 
interest.  How  many  days  are  there  in  1  yr.  5  mo.  17  da.  ?  527. 

204  x  Tfa  x  ,iT  x  527 

"Multiplying  $204  by  T^  gives  the  interest  for  1  yr.  at  7%  ;  divid- 
ing by  360,  for  1  da. ;  multiplying  by  527,  for  527  da.  Finish  the 
work;  in  cancelling,  never  divide  100  by  a  common  factor.  Why 
not  ?  The  interest  is  $20.90." 

Six  Per  Cent  by  Days.  To  find  the  interest  at  6%, 
move  the  decimal  point  of  the  principal  3  places  to  the 
left,  multiply  by  the  number  of  days  arid  divide  by  6. 
Modify  the  result  for  a  different  rate. 

ILL.  T.  "  Let  us  find  a  rule  for  computing  interest  at  6%.  Using 
the  cancellation  method,  find  the  interest  of  P  dollars  for  D  days  at 

6%. 

P  x^x-g^xD  =  .001  P  x  D  x  i 

"  Who  can  give  me  the  rule  ?  By  this  rule  what  is  the  interest  of 
$204  for  1  yr.  5  mo.  17  da.  at  6%  ?  $17.918.  At  7%?  $20.90. 

17.918        6 

2.986        1 

20.904        7 


§207  LESSON  29.    INTEREST  155 

"What  is  the  best  way  of  finding  interest  at  7%  from  interest  at 
6%?  7  is  1  more  than  6;  divide  by  6  and  add.  In  dividing  17.918 
by  6  the  exact  quotient  is  2.986 J.  Is  it  necessary  to  write  £  ?  Why 
not?  In  case  of  2.986f  what  would  we  have  done?  Written  7  in 
place  of  6.  Why? 

"  Give  me  the  complete  rule." 

Six  Per  Cent  Basis  SI.  Find  the  interest  of  81  for  the 
given  time  at  6  %  and  multiply  by  the  number  of  dollars. 
After  the  multiplication,  modify  the  result  for  a  different 
rate.  To  find  the  interest  of  $1,  count  the  interest  for 
1  yr.  6^;  for  1  month,  ^ff ;  for  1  day,  ^  m. 

ILL.  T.  "  Let  us  compute  interest  by  first  finding  the  interest  of 
$1  at  6%. 

"What  is  the  interest  of  $1  for  1  yr.  at  6%?  Gf<.  For  1  mo.? 
\t  (^  of  6f).  For  1  da.?  \  of  a  mill  (-&  of  5  m.).  Memorize  these 
facts. 

"  Find  the  interest  of  $  1  for  1  yr.  5  mo.  17  da.  at  6%.    See  next  page. 

"  The  interest  of  f  1  for  1  yr.  at  6  %  is  $  .06 ;  for  5  mo.,  $  .025 ;  for 
17  da.,  $.002f ;  for  the  whole  time,  $  .087$. 

"If  the  interest  of  $1  is  $.087£,  what  is  the  interest  of  $204? 
$  17.918.  If  the  interest  at  6%  is  f  17.918,  what  is  the  interest  at  7  %  ? 
$20.90." 

Aliquot  Part  Method.  Find  the  interest  for  a  year  at 
the  given  rate  and  modify  the  result  for  the  years,  the 
months  and  the  days. 

ILL.  T.  "  Let  us  compute  interest  by  first  finding  the  interest  for 
1  year. 

"  What  is  the  interest  of  $204  for  1  yr.  at  7%?" 

14.28  1  yr. 

1.19  1  mo. 

5.95  5  mo. 

.039|  1  da. 

.674  17  da. 
20.904 


156 


LESSON  29.    INTEREST 


§207 


Bankers'  Method.  To  find  the  interest  for  60  days  at 
6  %,  move  the  decimal  point  of  the  principal  two  places  to 
the  left.  Modify  the  result  for  a  different  time  or  rate. 

ILL.  T.  "Notes  are  usually  drawn  for  30  da.,  60  da.,  90  da.,  or 
120  da.  Let  us  find  a  rule  for  computing  interest  for  60  da.  at  6%. 

P  x  rfo  x  T*T  x  60  =  .01  P 

"  Who  can  give  me  the  rule?  At  6%  what  is  the  interest  of  $225 
for  60 da.?  $2.25,  Of  $  250  for  90  da.  ?  $3.75  ($2.50,  $1.25,  $3.75). 
Of  $300  for  30  da.  ?  $1.50  ($3,  $1.50)." 


CANCELLATION 

P,  $204 

T,  1  yr.  5  mo.  ggQ 

17  da.  150 

R,7%  J7 

I,  ?  527 

204xTJoX3JBX527 

I,  $20.90     Ans. 


6  %  BY  DAYS 
P,  $204 

T,  1  yr.  5  mo. 

17  da, 
T?    7  o/ 

**l    '    10 


.204X 
527 


107.508 

17.918  6 

2.986  1 

20.904  7 


I,  $20.90   Ans. 


6%  BASIS  $1 

P,  $204  .06 

T,  1  yr.  5  mo.  .025 

17  da.  ^02| 
B,-7% 

T  ?  — — 

17.918       6 

2.986       1 
I,  $20.90      Ans.         20.904       7 


ALIQUOT  PARTS 

P,  $204 

T,  1  yr.  5  mo.  14.28    1  yr. 

17  da.  L19             J  mo- 

T>    7  o/  5.95    5  mo. 

J%7  /0  .0391            1  da. 

*»  '  .674  17  da. 


20.904 


).90    Ans. 


BANKERS'  METHOD 


P,  $287.25 
T,  92  da. 
R,6% 
I,? 

I,  $4.40     Ans. 


2.87x25  60 

1.43  62  30 

.0957  2 
4.4044 


BANKERS'  METHOD 

P,  $287.25 

T,  120  da.  2.87x25       60 

K,  4 1  %  5.74  50      120 

I,  ?  1.43  62          1J 


4.30  88 


I,  $4.31     Ans. 


§208  LESSON  29.    INTEREST  157 

Discussion.  The  methods  to  be  used  with  vhe  class 
must  be  decided  from  the  course  of  study  or  by  the  prin- 
cipal of  the  school.  It  is  important  that  pupils  shall  gain 
the  power  to  compute  interest  mentally.  The  cancellation 
method  alone  is  not  sufficient. 

ILL.  A  graduate  of  the  N.  Y.  T.  S.  for  Teachers  could  not  com- 
pute the  interest  of  $  100  for  1  yr.  at  5%  mentally.  Given  a  pencil, 
she  produced  the  following  : 


This  young  woman  learned  the  cancellation  method  in  the  elemen- 
tary school,  failed  to  master  any  other  in  the  training  school,  and  was 
helpless  without  a  pencil. 

208.  Note  with  Interest,  Continued. 

ILL.  T.  "  Take  note  No.  2.  What  was  the  interest  at  maturity? 
The  interest  of  $1100  for  92  da.  at  4%,  or  $11.24.  How  did  you  get 
it,  Mary?" 

M.  "The  int.  of  $1100  for  'l  yr.  at  4%  is  $44;  for  90  da.  or  \ 
yr.,  $11;  for  10  da.  or  £  of  90  da.,  $1.22;  for  2  da.  or  \  of  10  da., 
$.24;  for  92  da.,  $11.24." 

T.  "  How  much  must  Henry  Jones  pay  Oct.  1  ?  $1111.24.  Give 
me  the  complete  history  of  this  note." 

209.  Bank  Discount.     To  find  bank  discount,  compute 
the  interest  on  the  amount  at  maturity  for  the  term  of 
discount  at  the  rate  of  discount. 

ILL.  T.  "  Take  note  No.  2  again.  Let  us  suppose  that  on  Aug.  1 
Smith  needed  money  and  obtained  it  by  selling  this  note  to  the  bank 
or  getting  it  discounted.  He  wrote  his  name  across  its  back  or  in- 
dorsed it,  agreeing  thereby  to  pay  it  in  case  Jones  should  fail  to  do 
so,  and  gave  it  to  the  bank.  Indorse  the  note. 

"  How  much  did  the  bank  pay  for  the  note?  On  Oct.  1  it  will  re- 
ceive $1111.24.  On  Aug.  1  it  paid  $  1111.24  less  the  interest  at  6%  from 
Aug.  1  to  Oct.  1  ;  it  simply  deducted  the  interest  in  advance  or  the  bank 
discount.  Bank  discount  is  interest  on  the  amount  at  maturity  for  the 
term  of  discount  at  the  rate  of  discount.  Memorize  this  statement. 


158 


LESSON  29.    INTEREST 


§210 


"We  must  find  the  interest  of  $1111.24  for  61  da.  at  6%;  it  is 
$11.30.  What  is  the  amount  at  maturity  less  the  bank  discount  or 
the  proceeds  or  what  Smith  will  receive? 

"Complete  the  history  of  the  note.  On  Oct.  1,  Jones  pays  the 
bank  $1111.24,  the  bank  writes,  'Paid  Oct.  1,  1912,  The  Tenth  Na- 
tional Bank,  John  Doe,  Cashier,'  across  the  face  of  the  note  or  cancels 
it,  and  returns  it  to  Jones.  Cancel  note  No.  2. 

"  Take  note  No.  1.  Suppose  that  Smith  had  sold  this  note  to  the 
bank  on  Aug.  1,  how  much  would  he  have  received  for  it?" 


NOTE  No.  1 

Face,  $  1100 

Date  of  note,  July  1,  1912 

Date  of  dis,     Aug  1,  1912 

R  of  dis,  6% 

Note,  without  int 

Proceeds,  ? 


Amt  at  mat,  $1100.00 
B  dis,  $11.18 

Proceeds,        $1088.82     Ans. 


NOTE  No.  2 

Face,  $1100 

Date  of  note,  July  1,  1912 

Date  of  dis,     Aug  1,  1912 

R  of  dis,  6  % 

Note,  with  int,  4  % 

Proceeds,  ? 

Amt  at  mat,  $1111.24 
Bdis,  $11.30 

Proceeds,        $1099.94     Ans. 


210.  Exact  Interest.  Banks  of  discount  sometimes 
count  360  and  sometimes  365  da.  to  the  year,  according 
as  it  is  to  their  advantage.  Thus,  they  count  360  when 
they  collect  interest  and  365  when  they  pay  interest.  The 
government  counts  365.  Unless  otherwise  directed,  count 
360. 

ILL.  T.  "  Sometimes  it  is  necessary  to  find  exact  interest  or  inter- 
est counting  365  da.  to  a  year. 

"Find  the  exact  interest  of  $204  from  July  1,  1911,  to  Dec.  18, 
1912,  at  7%.  Can  we  count  the  time  as  365  +  150  +  17  days  ?  No !  In 
reckoning  365  da.  to  a  year  5  mo.  cannot  be  counted  as  150  da.  The 
time  is  365  +  170  days  or  535  da.  Use  the  cancellation  method." 


204  x 


x  535 


§211  LESSON  29.    INTEREST  159 

211.  Postal  Savings  System.     Teachers  should  secure 
from  a  post  office  the  pamphlet,  "  Postal  Information," 
should  explain  its  provisions,  should  give  problems  based 
upon  it,  should  exhibit  postal  savings  cards,  ten -cent  sav- 
ings stamps,  and  savings  certificates,  and   should  urge 
pupils  to  become  depositors. 

ILL.  T.  "  Have  any  of  you  deposited  money  in  the  postal  savings 
bank  at  the  post  office?  How  old  must  a  person  be  to  open  an  ac- 
count ?  10  jr.  How  much  can  you  deposit  at  one  time  ?  An  exact 
number  of  dollars ;  but  you  can  buy  10-cent  stamps  and  stick  them 
to  a  card,  and  when  you  have  a  dollar  in  stamps  you  can  make  a 
deposit.  Here  is  a  postal  savings  card  with  5  stamps.  How  much  is 
it  worth  ? 

"  When  you  make  a  deposit,  you  receive  a  savings  certificate  which 
bears  interest  at  2  %  for  every  full  year  the  money  is  on  deposit,  be- 
ginning with  the  first  day  of  the  month  following  the  one  in  which 
it  is  deposited.  A  person  has  as  many  certificates  as  he  makes  de- 
posits. Here  is  a  postal  savings  certificate. 

"Jan  2,  1912,  I  obtained  a  savings  certificate  of  $3 ;  Oct.  3,  1912, 
one  of  $  4 ;  and  Apr.  30,  1913,  one  of  <$  2.  How  much  are  these  cer- 
tificates worth  to-day,  counting  interest  ?  " 

212.  Exercises.     1.   Compute  the  interest  of  $511  for  11  mo.  11  da. 
at  5  %  by  the  cancellation  method.      2.   By  the  6  %  method  for  days. 
3.   By  the  6  %  basis  <$  1  method.      4.   By  the  aliquot  part  method. 
6.   Find  the  rule  for  computing  interest  at  36  %.      6.   By  this  rule 
compute  the   interest  in   No.    1.       7.   State,   with  reasons,   which 
method  you  prefer.       8.    Discuss  the  bankers'  method.      9.   Write 
note  No.  1,  making  the  time  90  days  after  date,  and  adding  '  with 
interest.'       10.    Find  the  proceeds  of  the  note  July  20,  discounting 
at  6  %.      11.   Prove  that  for  less  than  a  year  exact  interest  is  the  com- 
mon interest  minus       of  the  common  interest. 


LESSON   30.     INTEREST 

213.  Bonds.  A  note  given  by  a  village,  town,  city, 
county,  state,  or  nation,  secured  by  faith  and  credit,  and 
bearing  interest,  is  called  a  bond.  Bonds  are  usually 
treated  in  arithmetics  in  connection  with  stocks,  but  all 
which  they  have  in  common  with  stocks  is  the  fact  that 
they  are  usually  bought  and  sold  by  brokers.  Notes 
given  by  individuals  and  corporations  secured  by  mort- 
gage and  bearing  interest  are  also  called  bonds. 

ILL.  T.  "  Suppose  the  city  of  Yonkers  needed  $  500,000  on  June  1, 
1912,  to  construct  a  sewer,  and  the  money  in  the  treasury  was  in- 
sufficient, it  might  issue  and  sell  to  the  highest  bidder  500  notes 
(bonds)  of  $  1000  each,  signed  by  the  city  officers  as  provided  by  law. 
The  bonds  in  blank  might  read  : 


CITY  OF  YONKERS  SEWER  LOAN,  No.  _ 

YONKERS,  N.  Y.,  fane,  I,  19  IS. 
$/  00  0.00 

<Hn    fane,   I  ,    /tf32,    to^   vatuz   i&efd,    the-   ^-ity  of- 


MAe*  to- 


and  —  jbottoA*,  at 

100 


at  */•  %  foa/ucd>te>  o-n  fan&  /  at  &a&ti  yza-v  at 
-id  (mn-k.       &ti&  fattA,  and  vucUt  o^-  the*   @-tty  o^ 
to-  the*      vuwb&nt  at  tAC&  d&kt. 


.MAYOR 


_PRE8.  OF  COUNCIL  CITY  TREASURER 

160 


§214  LESSON  30.    INTEREST  161 

"  For  the  convenience  of  the  purchasers,  twenty  small  notes,  called 
coupons  (French,  cut),  might  be  printed  on  the  same  paper,  each  for 
f  40,  or  the  payment  of  one  year's  interest,  and  made  payable  in  order, 
June  1,  1913,  June  1,  1914)  and  so  on.  In  this  case  the  bond  would 
be  a  coupon  bond.  The  coupons  might  be  arranged  as  follows : 


No. 

20 

No. 

19 

No. 

13 

No. 

IT 

No. 

16 

No. 

15 

No. 

14 

No. 

13 

No. 

12 

No. 

11 

No. 

10 

No. 

9 

No. 

8 

No. 

T 

No. 

6 

No. 

5 

No. 

4 

No. 

3 

No. 

2 

No. 

1 

"  Suppose  this  was  a  coupon  bond.  Write  coupon  No.  1,  or  the 
note  becoming  due  June  1,  1913.  Explain  the  procedure  of  Mr.  A., 
who  owns  one  of  these  bonds,  in  getting  his  interest  each  year." 

Bond  and  Mortgage.  A  person  owning  real  estate  may 
borrow  money  by  giving  to  the  lender  a  note  called  a 
bond  (bound)  and  a  sealed  instrument  in  writing  trans- 
ferring the  property  to  him.  This  instrument  is  a  deed  or 
mortgage  (death-pledge).  The  lender  has  this  deed  re- 
corded to  prevent  fraud,  but  is  in  no  sense  the  owner  of 
the  property.  If  the  borrower  fails  to  fulfill  his  agree- 
ment, the  property  is  sold  under  the  direction  of  the  court 
and  the  lender  is  paid. 

214.  Drafts.  If  A  owes  B,  B  may  request  A  to  pay 
the  whole  or  a  part  to  C.  If  A  honors  the  request,  B  may 
pay  C  in  this  way.  The  paper  containing  the  request  is 
a  draft.  It  may  or  may  not  bear  interest  and  may  be 
payable  on  demand  or  at  a  specified  time. 

ILL.  T.  "  George  Williams  owes  the  Home  Publishing  Co.  $  1000 
for  books.  The  latter  makes  a  draft  on  the  former  for  $  500  in  favor 
of  Henry  Brown,  payable  30  da.  after  sight. 


162  LESSON  30.    INTEREST  §215 


To  ^WME,  l&LttLa/m&f  300  BROADWAY,  N.  Y.  CITY, 

220  BwadwiMf,  cA.  If.  &lty.  WOA,.  25,  iq/3. 

a/u  to  tA.&  o-'uie.^  o 


am-cl  — 

100 

at  want. 


f  PRES. 


"  The  draft  is  sent  to  Williams,  who  writes  across  the  face  in  red 
ink,  '  Accepted,  Mar.  26,  1913,  George  Williams.'  It  is  then  sent  to 
Henry  Brown,  who  sells  it  or  collects  it  when  due.  Write  the  draft 
and  its  acceptance." 

Checks.  The  most  common  form  of  draft  is  a  check. 
A  person  has  money  on  deposit  in  bank  and  pays  debts 
by  checks.  As  a  rule,  checks  are  not  sent  to  the  bank 
for  acceptance,  but  sometimes  this  is  necessary  when  the 
validity  of  the  check  is  to  be  put  beyond  question.  In 
such  an  event  the  check  is  said  to  be  certified.  It  be- 
comes a  draft  which  has  been  accepted  by  the  bank.  A 
bank  officer  writes  the  acceptance  in  the  usual  way. 

ILL.  T.  "  We  will  suppose  that  you  have  money  on  deposit  at  the 
Tenth  National  Bank,  100  Broadway,  N.  Y.  City.  Draw  a  sight 
draft  on  the  bank  for  no  dollars  to  the  order  of  James  Thompson. 
This  draft  is  a  check. 

"  Have  this  check  certified.  That  is,  get  one  of  your  classmates  to 
accept  it  as  cashier  in  the  name  of  the  bank." 

215.  Indirect  Cases.  The  direct  cases  have  already 
been  considered  ;  given  the  principal  time  and  rate  to  find 
the  interest,  and  to  find  the  amount,  that  is  the  principal 
plus  the  interest.  They  should  receive  90%  of  the  time 
devoted  to  interest. 


5215  LESSON  30.    INTEREST  163 

The  indirect  cases  grow  out  of  the  direct  cases.  Six 
of  them  are  usually  considered  in  the  arithmetics. 

ILL.  T.  "The  interest  of  $200  for  2  yr.  at  6%  is  $24.  We  have 
mastered  this  case.  Find  the  three  others  that  arise  from  the  omis- 
sion of  each  term  in  succession." 

1.  In  what  time  will  $200  gain  $24  at  6%? 

2.  At  what  rate  will  $  200  gain  $  24  in  2  yr.  ? 

3.  What  principal  will  gain  $24  in  2  yr.  at  6%? 

"The  amount  of  $200  for  2  yr.  at  6%  is  $224.  Find  the  three 
other  cases." 

4.  In  what  time  will  $200  amount  to  $224  at  6%? 

5.  At  what  rate  will  $200  amount  to  $224  in  2  yr.? 

6.  What  principal  will  amount  to  $224  in  2  yr.  at  6%? 

By  Analysis.  The  author  prefers  this  method.  The 
component  problems  are  displayed  in  the  diagrams.  See 


/En 

'•  T\  , 

I,  $24 
/P,  $200 

yr.<T 

\R,6% 

/A,  $224 
Enl< 
4    T/         \P,  $200 
\               /P,  $200 
-  1  yr./ 

XR,  6% 

<En 
I  1 

I,  $24 
/P,  $200 
%< 
\T,2yr. 

/A,  $224 
Enl< 
A  R/         \P,  $200 
\             /P,  $200 

XT,  2  yr. 

/En 

I,  $24 

/En  A,  $224 

I  $ 

l<f 

61    P\           /T,  2  yr. 

\R    R% 

A  $  1^ 

•*•*>  "  /O 

\R,6% 

ILL.  T.  "Consider  Case  1.  From  what  can  we  find  the  no.  of 
years?  From  the  entire  interest  and  the  interest  for  1  yr.  From 
what  can  we  find  the  int.  for  1  yr.  ?  From  the  principal  and  the  rate." 

ILL.  T.  «  Solve  Case  1.  What  is  the  interest  of  $  200  for  1  yr.  at 
6  %  ?  $  12.  If  the  interest  is  $  12  for  1  yr.,  in  how  many  years  will  it 
be  $24?  2. 


164 


LESSON  30.    INTEREST 


§216 


"  Solve  No.  5.  If  the  amount  is  $  224  and  the  principal  is  $2( 
what  is  the  interest  ?  $  24.  What  is  the  interest  of  $  200  for  2  yr.  at 
1  %?  $  4.  If  the  interest  is  $  4  at  1  %,  at  how  many  per  cent  will  it  be 
$  24  ?  6." 

By  Algebra.  The  known  terms  may  be  substituted  in 
the  direct  formulse,  or  the  unknown  term  may  be  repre- 
sented by  x.  See  p.  IJ+l. 


IST  ALGEBRA 

2u  ALGEBKA 

P,  $200 
A,  $224 

I=200x2.X  R 
224  =  200  +  1 

P,  $200 
A,  $224 

Let  —  =  R 
100 

T,  2  yr. 

J=24 

T,  2  yr. 

200  X  2  x  — 
100 

R,? 

24  =  200  X  2  X  R 

R,?        • 

•-T 

P                24                am 

w 

R,6% 

Ans. 

R,  6% 

A  tft  O                                                                  fa,    R 
^L  /*O.                                                                J£,    \J 

FORMULA 

ANALYSIS 

P,  $200 

1=  Px  Tx  R 

P,  $200 

A,  $224 

A  =  P  +  I 

A,  $224 

224 

T,  2  yr. 
R,? 

R-      J 

T,  2  yr. 
R,? 

200 
24 
200  X  .02  = 

4 

Px  T 
I=A-P 

R,  6% 

R     A  —  P      24 

R,  6% 

24  -r  4  = 
Ans. 

6 

Ans.                Px  T     400 

216.  Different  Cases.  For  the  Teacher.  Let  us  find  all 
the  different  cases  in  interest  from  the  formulae,  1=  P  x 
T  x  R  and  A  =  P  +  I.  See  §  7. 

There  are  two  equations  with  five  quantities.  To  solve  these 
equations  three  of  the  quantities  must  be  known.  Hence,  for  every 
three  known  terms  there  will  be  two  cases. "! 

The  combinations  of  ttavixMn  the  terms,  A,  I,  P,  R,  T  are  A  IP, 
AIR,  AIT;  APR,  APT;  ART;  IPR,  IPT;  IRT;  PRT.  That  is, 
there  are  10  times  2,  or  20,  cases  in  interest. 

Write  the  20  cases.     Two  of  them  are  impossible.     Why? 


§217 


LESSON  30.     INTEREST 


165 


217.    Kinds  of  Interest.     For  the  Teacher.     Let  us  clas- 
sify interest  with  reference  to  what  may  bear  interest. 
ILL.     What  is  the  interest  of  $  100  for  4  yr.  @  6  %? 


i 

IST  YE.            2D  TK.           3D  YB. 

4TH  YR 

$6               $6               $6 

$6 

I  on  P,  $24 

.36              .36 

.36 

.36 

.36 

I  on  I  on  P  at  the  end 

.36 

of  each  year,  $2.16 

.0216 

.0216 

.0216 

I  on  all  other  unpaid 

.0216 

I  at  the  end  of  each 

.001296 

year,  $  .087696 

First  Conception.  The  principal  alone  may  bear  interest,  simple 
interest  ($  24). 

Second  Conception.  In  addition  to  the  above,  the  $  6  due  at  the 
end  of  the  1st  year  may  bear  interest  for  3  yr. ;  the  $  6  due  at  the  end 
of  the  2d  year  may  bear  interest  for  2  yr. ;  the  $  6  due  at  the  end  of 
the  3rd  year  may  bear  interest  for  1  yr.  That  is,  the  principal  may 
bear  interest  and  the  interest  on  the  principal  at  the  end  of  each  year 
may  bear  interest,  annual  interest  ($26.16). 

Third  Conception.  In  addition  to  the  above,  the  $  .36  due  at  the 
end  of  the  2d  year  may  bear  interest  for  2  yr. ;  each  $  .36  due  at  the 
end  of  the  3d  year  may  bear  interest  for  1  yr. ;  the  $  .0216  due  at 
the  end  of  the  3d  year  may  bear  interest  for  1  yr.  That  is,  the  prin- 
cipal may  bear  interest,  the'interest  on  the  principal  at  the  end  of  each 
year  may  bear  interest  and  all  other  unpaid  interest  at  the  end  of 
each  year  may  bear  interest,  compound  interest  ($26.247696). 

218.  Annual  Interest.  For  the  Teacher.  Annual  inter- 
est is  rarely  computed.  The  above  development  indicates 
the  method  of  procedure. 

ILL.  What  is  the  annual  interest  of  $  525.26  for  3  yr.  5  mo.  17 
da.  <g  6? 


166  LESSON  30.    INTEREST  §219 

P,  f  525.26 

T,  3  yr.  5  mo.  17  da.  5.25x26        2    5    17 

R,6  6        1    5    17   . 

An  I,?  31-5156  5    17 
4    4    21 

I  on  P,  $  109.17  $525.26  bears  interest  3  yr.  5  mo.  17  da. 

I  on  I,       $  8.30  f  31.5156  bears  interest  4  yr.  4  mo.  21  da. 
An  I,     $  117.47  'Ans. 

219.  Compound  Interest.  For  the  Teacher.  Compound 
interest  is  rarely  computed.  For  practical  use,  the  third 
conception  is  reduced  to  the  form:  in  compound  interest 
the  amount  at  the  end  of  each  year  bears  interest  during 
the  ensuing  year. 

ILL.     What  is  the  amount  of  $  100  for  4  yr.  6  mo.  @  6  %? 

P,  $100 

T,  4  yr.  6  mo.  106  1        119.1016 

R  6°/  1.06                   1.06 

Com  I,?  112M  2        126.247696    4 

1 .          1.06  1.03 

A,  $  130.04  119.1016  3        130.035126 
"  Com  I,  $30.04  .4ns. 

EXPL.  The  amount  at  the  end  of  the  4th  year  is  $  126.247696 ; 
the  amount  of  $  1  for  6  mo.  is  $  1.03  ;  the  amount  of  $  126  +  is  126  + 
times  $  1.03,  etc. 

Second  Plan.  The  above  is  the  popular  method.  A 
better  plan  is  to  find  the  amount  of  $  1,  and  to  multiply 
by  the  number  of  dollars.  Before  multiplying  it  is  well  to 
express  the  amount  of  the  principal  by  its  factors. 

ILL.  What  is  the  amount  of  $  525  for  20  yr.  7  mo.  17  da.  @  6  % 
compound  interest? 

P,  $525 

T,  20  yr.  7  mo.  17  da.  .035 

R,  6%  ^°2| 

OS7* 

/0m      '  A  =  525  x  1.0620  X  1.0371 

Com  A,  $1747.43  Ans. 


§220  LESSON  30.    INTEREST  167 

EXPL.  The  amount  of  $  1  at  the  end  of  the  1st  yr.  is  $  1.06  ;  at  the 
end  of  the  2d,  $  1.062;  at  the  end  of  the  3d,  f  1.068;  ...  at  the  end 
of  the  20th,  $  1.0620  ;  the  amount  of  f  1  for  7  mo.  17  da.  is  f  1.037| ; 
the  amount  of  $1  for  the  whole  time,  ftl.OG20  x  1.037$. 

The  value  of  1.0620  can  be  found  by  multiplication  as  in  §  178, 
from  tables,  or  by  the  use  of  logarithms. 

Interest  Periods.  When  interest  is  payable  semi- 
annually  or  quarterly,  it  is  well  to  change  the  problem  to 
an  equivalent  problem  in  which  the  interest  is  payable 
annually.  It  is  evident  that  each  interest  period  may  be 
regarded  as  a  year  if  the  rate  is  divided  by  the  number  of 
such  periods  in  a  year. 

ILL.  Show  the  amount  of  $  525  for  20  yr.  7  mo.  17  da.  at  6  %  in- 
terest compounded  semiannually.  Ans.  A  =  525  x  1.0341  x  1.007|. 

This  means,  what  is  the  amount  of  $  525  for  41  yr.  3  mo.  4  da. 
(twice  20  yr.  7  mo.  17  da.)  at  3  %  (half  of  6  %)? 

Discussion.  In  all  states,  the  laws  are  opposed  to  the 
collection  of  compound  interest.  The  difference  between 
the  amount  at  simple  and  at  compound  interest  is  well 
illustrated  by  computing  the  interest  of  1  r  at  6  %  by  each 
method  since  the  beginning  of  the  Christian  era.  For 
1913  years,  the  interest  by  one  method  is  $  1.15  ;  by  the 
other,  many  times  the  value  of  a  sphere  of  pure  gold 
whosev radius  is  the  distance  from  the  earth  to  the  sun. 

220.  Savings  Banks.  Savings  banks  pay  a  modified 
form  of  compound  interest.  At  the  end  of  each  interest 
period,  the  interest  of  the  largest  sum  that  has  been  on 
deposit  during  the  entire  period  is  added  to  the  amount 
of  the  deposit,  but  interest  is  computed  on  an  integral 
number  of  dollars  only. 

ILL.  T.  "  Have  any  of  you  deposits  in  a  savings  bank?  Be  pre- 
pared to  tell  me  to-morrow  the  rule  for  computing  interest  stated  in 
your  bankbook.  What  did  you  find,  James?" 


168  LESSON  30.    INTEREST  §221 

J.  "  My  account  is  in  the  People's  Savings  Bank  of  Yonkers.  The 
statement  is,  '  On  the  first  day  of  Jan.  and  July  of  every  year, 
there  shall  be  declared  and  paid  interest  on  all  sums  of  $5  and  up- 
wards, which  shall  have  been  deposited  for  3  mo.  previous  to  the  first 
day  of  Jan.  and  July;  but  no  interest  shall  be  paid  on  the  fractional 
part  of  a  dollar,  nor  shall  any  interest  be  allowed  on  any  sum  with- 
drawn previous  to  the  first  day  of  Jan.  and  July  for  the  period  which 
may  have  elapsed  since  the  last  dividend.'  Jan.  1,  my  father  opened 
an  account  for  me  with  $  10 ;  July  1,  I  shall  have  $  10.20,  and  next 
Jan.  1,  $  10.40." 

He  develops  the  subject  at  length  with  an  imaginary  account. 

"  Henry  Brown  opened  an  account  with  the  People's  Savings  Bank, 
Jan.  1,  1912,  by  depositing  |150.98;  Mar.  10,  he  withdrew  $50; 
Apr.  1,  he  deposited  $100.40.  How  much  did  he  have  on  deposit 
July  1,  1912,  after  interest  @  4  %  was  added? 

His  account  appears  as  follows  : 


DEPOSITS                       DRAFTS  BALANCE 

9(M>1.    I                    loO.tfS  /50.<?8 

/q/2,  TWO*,,  /o                           30.00  too.qs 

/<?I2,  ftfa,.  /              /OO.V-O  20/.38 
f<// 2,  fuly.  I    •               3. Of  (M.) 


"  The  largest  whole  no.  of  dollars  on  deposit  during  the  entire  six 
months  is  $  100  ;  it  bears  $1  interest  during  the  1st  quarter  (1  %  of 
$  100)  ;  the  largest  whole  no.  of  dollars  on  deposit  during  the  entire 
2d  quarter  is  $201  ;  it  bears  $2.01  interest  during  the  2d  quarter 
(1%  of  $201)." 

221.  Exercises.  1.  Read  the  diagrams  in  §  215.  Thus,  Case  2,  the 
rate  can  be  found  from  the  entire  interest  ($24),  and  the  interest  at 
1  %.  2.  Solve  Case  2  by  analysis.  3.  Solve  Case  3  by  analysis. 
4.  Solve  Case  4  by  analysis.  6.  Solve  Case  6  by  1st  algebra  method. 
6.  By  2d.  7.  By  formula.  8.  By  analysis.  9.  State,  with  rea- 
sons, the  method  you  prefer.  10.  Verify  the  answer  to  the  first 
problem  in  §  219. 


PAtfT  III.     EXERCISES 

SECTION  1.     ELEMENTARY  SCHOOLS 

222.  Scope.     This  section  gives  an  idea  of  the  advance 
in  arithmetic  which  is  made  from  term  to  term  in  the 
elementary  schools  of  New  York  City.     The  exercises  for 
the  first  two  years  have  been  suggested  by  the  author. 
The  exercises  for   the  last  six  years,  with   some   slight 
changes,  have  been  given  as  final  tests  by  those  in  authority. 

In   mental  tests,  the  questions  were  dictated   by  the 
teacher  and  answers  only  were  written  by  the  pupil. 

FIRST  TEAR 

223.  ist  Yr.  ist  T.     1.    On  your  desk  there  is  a  package  of 
sticks  and  some  rubber  bands.     Put  them  into  bundles  of  ten 
and  tell  me  how  many  sticks  you  have.       2.    Bring  me  thirty- 
four  sticks.       3.    Eead  these  numbers :  25,  30,  87,  17.      4.   By 
figures,  write :  7,  2,  4.       6.    How  far  is  it  from  your  desk  to 
mine    (answer  in  steps)  ?       6.   Find  the  answer  by  counting 
objects ;  there  are  sticks  and  pieces  of  paper  on  your  desks. 
Jane  has  4  dolls  and  her  mother  gives  her  2  more ;  how  many 

169 


170  SECTION    1.    ELEMENTARY  SCHOOLS  §224 

has  she  then  (Lesson  6)  ?  7.  As  before,  John  has  2^  and 
wants  to  buy  a  top  for  5  X ;  how  many  cents  does  he  lack  ? 
8.  As  before,  Mary  has  4  dolls  and  2  dresses  for  each ;  how 
many  doll-dresses  has  she  ?  9.  As  before,  if  there  are  2  plants 
in  a  row,  how  many  rows  will  10  plants  require  ?  10.  As 
before,  the  teacher  gives  8  books  to  4  girls,  to  each  one  the 
same  number ;  how  many  books  does  each  girl  receive  ? 

224.  ist  Yr.  2d  T.     1.    From  my  desk  get  bundles  of  ten 
and  single  sticks  and  express  forty-seven  by  objects.      2.  Write 
47  by  figures.      3.  Write  76  by  figures.      4.  Write  20  by  figures. 
5.    Take  20  sticks  from  this  box,  2  at  a  time.       6.    I  am  going 
to  buy  things  of  you  and  ask  you  to  make  change  with  the  toy 
money  on  your  desks.     I  buy  a  top  for  4  ^  and  pay  with  a  dime 
(p.  72) ;  make  the  change.      7.  Add :  3,  4,  2 ;  8,  2,  3.      8.  Draw 
the  objects  and  find  the  answer.     How  much  will  6  oranges 
cost  at  3^  each  ?      9.   As  before,  4  oranges  were  put  on  a  plate ; 
how  many  plates  were  needed  for  12  oranges  ?      10.  As  before, 
10  hat  pins  were  used  in  5  hats ;  each  hat  had  the  same  num- 
ber of  pins ;  how  many  pins  in  each  hat  ? 

SECOND  YEAR 

225.  2d  Yr.  ist  T.     1.    On  your  desks  there  are  bundles  of 
ten,  single  sticks,  and  bands.     Put  the  tens  into  bundles  of  a 
hundred  and  tell  me  how  many  sticks  you  have.       2.   Bring 
me  321  sticks.       3.   Number  to  8  and  remember  your  number. 
The  first  eight  may  take  places  at  the  board  ;  draw  a  circle ; 
write  in  the   center  a  number   one   greater   than  your   own; 
around  the  circle  write  the  digits  through  9  in  miscellaneous 
order;  be  seated.      4.    John,  call  all  the  sums  on  the  board 
keeping  in  time  with  the  sound  (he  taps  at  the  rate  of  two  com- 
binations a  second  and  calls  upon  another  if  John  fails  at  any 
point).       6.    Add:    379,    539.       6.   From   678   subtract  253. 
7.   Each  of  6  boys  has  4  marbles.     Make  a  mark  for  each  boy 
and  write  4  under  each.    How  many  marbles  in  all  ?      8.  Erase 


§226  SECTION   1.    ELEMENTARY  SCHOOLS  171 

the  4  under  each  mark.  Solve  No.  7  again.  9.  There  are  20 
crayons  and  some  boxes ;  each  box  has  4  crayons ;  how  many 
boxes  ?  Count  to  20  by  4's,  making  a  mark  at  each  count. 
10.  If  I  divide  18  little  cakes  among  6  boys,  giving  to  each 
boy  the  same  number,  how  many  cakes  will  each  boy  get? 

226.  2d  Yr.  2d  T.     1.   Count  from  1  to  10  as  fast  as  possible. 
2.    At  the  same  rate  count  from  1  to  97  by  4's.      3.  From  1  to 
100  by  3's.       4.    From  1  to  99  by  2's.      6.    Add :  394,  87,  109. 

6.  From  925  subtract  288.      7.    Draw  a  line  and  mark  off  f  of 
it.      8.   At  3  for  5^  how  much  will  12  bananas  cost?     Make  a 
mark  for  each  group  of  3  bananas ;  it  will  also  stand  for  each 
group  of  5  ^.      9.    At  4  for  5^  how  many  peaches  can  I  buy  for 
25^?     Make  a  mark  for  each  5  ^.      10.    At  21  ^  each  how  much 
will  4  books  cost  ?     Write  21  4  times  and  add. 

THIRD  YEAR 

227.  3d  Yr.    ist  T.     Mental.     1.   Spent  8^,   5#,  9^,   4^. 
How  much  was  spent?       2.   Paid  32^  for  4  Ib.  of  tomatoes. 
What  was  the  cost  of  1  Ib.  ?      3.    If  a  quart  of  milk  costs  9^, 
what  will  a  gallon  cost  ?      4.    How  many  legs  have  eight  flies  ? 

5.  I  have  one  yard  of  cloth.     If  I  cut  off  10  inches,  how  many 
inches  will  be  left  ?       6.    20  weeks  are  how  many  months  ? 

7.  I  had  8  dimes  and  spent  half  a  dollar.     How  much  money 
had  I  left  ? 

Written.     1.    Add :    2701,  1976,  2000,  3006.      2.  From  9654 
take  7235.       3.   Multiply  642  by  26.       4.   Divide  664  by  4. 

6.  Find  ^  of   924.       6.    How  many  gallon  measures  will  96 
quarts  of  milk  fill  ?      7.   In  234  feet  there  are  how  many  yards  ? 

8.  A  boy  was  on  board  a  steamer  1£  days.     How  many  hours 
was  he  on  board  ? 

228.  3d  Yr.  2d  T.     Mental.     1.    Write  the  Roman  number 
for  99.       2.   A  farmer  has  37  cows.     He  buys  62  more.     How 
many  has  he  then  ?       3.   If  3  oranges  cost  5  ^,  what  will  12 


172  SECTION   1.    ELEMENTARY  SCHOOLS  §229 

oranges  cost  ?  4.  Louis  has  36  marbles.  His  brother  has  f 
as  many.  How  many  has  his  brother  ?  6.  John  bought  30^ 
worth  of  cake  and  gave  the  baker  a  half  of  a  dollar.  How 
much  change  was  due  ?  6.  There  are  48  boys  in  the  3-B  class. 
If  f  of  them  are  promoted,  how  many  boys  are  promoted  ? 

7.  Change  56  days  to  weeks.      8.    If  6  handkerchiefs  cost  36  ^, 
what  will  7  handkerchiefs  cost  ? 

Written.  1.  Write  and  add:  $365.05,  $409.30,  $50.75, 
$  800.28,  $  749.06,  $  23.53.  2.  A  man  buys  a  farm  for  $  6550 
and  sells  it  for  $9375.  Did  he  gain  or  lose  and  how  much? 
3.  If  six  suits  of  clothes  cost  $  29.88,  what  is  the  cost  of  nine 
suits  ?  4.  Which  is  larger,  ^  or  ^  ?  Diagram.  6.  Show  that 
f  =  |.  Diagram.  6.  Multiply:  $96 .54  by  59.  7.  A  mer- 
chant sold  95  yards  of  cloth  to  one  man,  117  yards  to  another, 
and  then  had  49  yards  left.  How  many  yards  had  he  at  first  ? 

8.  At  18^  per  dozen,  what  is  the  cost  of  16  pencils  ?      9.  How 
many  5ths  in  23  units  ? 

FOURTH  YF.AE 

229.  4th  Yr.  ist  T.  Mental.  1.  7,  x  8,  +4,  -  10,  -=-  5, 
—  1,  X  9  ?  2.  Had  56  children  in  a  class ;  promoted  £  of 
them.  How  many  were  promoted  ?  3.  Spent  3  /  for  a  pencil, 
5^  for  a  book,  and  2^  for  a  pen.  What  change  did  I  receive 
from  half  a  dollar  ?  4.  How  many  ounces  in  10  Ib.  of  tea  ? 
6.  If  5  books  cost  $  1.75,  what  will  1  book  cost  ? 

Written.  1.  38704 --59  =  What?  2.  6041  x  807=  What? 
3.  If  25  suits  cost  $  625,  what  will  1  suit  cost  ?  4.  Sold  a 
house  for  $4500  and  lost  $800.  What  was  the  cost?  6.  A 
man  earns  $1072  a  year;  he  spends  |  of  it.  How  much  does 
he  save  ?  6.  How  many  days  in  100  years  if  75  of  them  have 
365  da.  each,  and  25  have  366  da.  each  ?  7.  What  will  5f  lots 
cost  if  one  lot  costs  $164  ?  8.  What  is  the  cost  of  40  quarts 
of  milk  at  2  cents  a  pint?  9.  A  has  $25284  and  B  has 
$87611.  How  much  more  money  has  B  than  A?  10.  Add: 


§230  SECTION   1.    ELEMENTARY  SCHOOLS  173 

two  hundred  seventy-five,  three  hundred  eighty-four,. twenty 
six,  nine,  five  thousand  two  hundred  eighty-four,  six  hundred 
fifty-eight,  ninety-seven,  five  thousand  two  hundred  sixty- 
four. 

230.  4th  jr.  2d  T.      Mental    1.   If    a   yard  of    silk  costs 
$  1.25,  what  will  8  yards  cost  ?       2.    What  part  of  a  yard  is 
18  inches  ?         3.    Mr.  Jones  had  36  sheep.      Wolves  killed 
£  of  them.      How  many  were  left  ?         4.    What  is  the  num- 
ber of  square  feet  in  the  floor  of  a  room  11  ft.  by  12  ft.  ? 

5.  Bought  books  for  $25.     Sold  them  and  gained  -|  of  the 
cost.      What  was  the  selling  price?         6.    A  square  field  is 
10  rods  on  a  side.     How  far  around  it  ?       7.   What  is  the 
value  of  a  blackboard  4  feet  long  by  5  feet  wide,  at  50  cents  a 
square  foot?         8.   At  25  cents  a  peck,  what  is  the  cost  of 
2  bushels  of  potatoes  ?         9.   How  many  square  inches  in  a 
3-inch  square  ?         10.    Reduce  ^  to  a  mixed  no. 

Written.  1.  Add :  879,  789,  897,  688,  987,  438.  2.  From 
nine  hundred  thousand  eight  hundred  six,  subtract  eight  hun- 
dred forty-five  thousand  nine  hundred  eighty-seven.  3.  Had 
$  752 ;  spent  |  of  it ;  how  much  was  left  ?  4.  If  360  books 
cost  $  900,  what  is  the  cost  of  1  book  ?  6.  Find  the  cost  of 
fencing  a  lot  100  ft.  long  and  50  ft.  wide  at  50  cents  a  foot. 

6.  How  many  lots  25  x  100  can  be  made  from  a  plot  75  x  200  ? 

7.  Bought  a  piano  for  $  350  and  sold  it  at  a  loss  of  $  75.50 ; 
find  the  selling  price.       8.    A  man  earned  S  900  a  year,  spent 
$750  each  year,  and  saved  the  remainder.     How  much  was 
saved  in  10  years  ?        9.    Henry  weighs  58f  lb.,  Peter,  65f  lb., 
and  John,  67|  lb.     Their  father  weighs  as  much  as  all  three 
together.     What  is  his  weight  ?       10.    How  many  cubic  feet 
of  earth  must  be  removed  to  make  a  cellar  36  ft.  by  28  ft.  by 
14  ft.  ? 

FIFTH  TEAR 

231.  sth  yr.  ist  T.     Mental     1.   Add:    15,  9,  12,  6,  8,  9. 
2.   James  has  $2£  and  Harry  $5^.      How  much  have  they 


174  SECTION   1.    ELEMENTARY  SCHOOLS  §232 

together  ?  3.  If  a  car  goes  4  miles  in  half  an  hour,  how  far 
will  it  go  in  3^  hours  ?  4.  A  man's  salary  is  $  24  a  week. 
He  spends  $  15  each  week.  What  part  of  his  salary  does  he 
save  ?  6.  John  had  20  marbles,  and  James  f  as  many. 
How  many  did  both  have  ?  6.  If  I  burn  f  of  a  ton  of 
coal  in  a  month,  how  long  will  6  tons  last  ?  7.  2|-  pt.  are 
what  part  of  5  qt.  ?  8.  Bought  2  bbl.  of  apples,  one  for 
$  1.25,  the  other  for  $  1.50.  What  change  did  I  receive  from 
a  $  5  bill  ?  9.  A  family  use  6  eggs  each  morning.  How 
many  eggs  do  they  use  in  the  month  of  February  of  this  year  ? 
10.  John  paid  $  16  for  his  overcoat,  which  was  -|  of  the  cost  of 
his  suit.  What  did  he  pay  for  his  suit  ? 

Written.  1.  Paid  $  800  for  a  plot  of  land  200  feet  long  and 
100  feet  wide,  which  I  divided  into  lots  25  ft.  x  100  ft.,  and 
sold  for  $  150  each.  How  much  did  I  gain?  2.  I  had 
75^-  acres  of  land  and  sold  18f  acres  to  one  man,  3f  acres  to  a 
second,  and  5-f  acres  to  a  third.  How  many  acres  had  I  left  ? 

3.  If  1  yard  of  cloth  cost  $  3|,  find  the  cost  of  18f  yards. 

4.  Make  out  and  receipt  the  following  bill,  your  mother  the 
purchaser,  the  date  to-day.      Bought  of  E.   H.  Macy  &  Co., 
34th  St.  and  Broadway,  12|  yd.  lace  @  $  1.50 ;   20  yd.  linings 
@12£l;   24  yd.  ribbon  @  7£^;  15  spools  of  thread  @  9  t. 

5.  From  a  crop  of  324  bu.  a  man  sold  243  bu.     What  part  of 
his  crop  had  he  left  ?       6.   If  £  of  a  farm  is  valued  at  $  86.24, 
what  is  the  value  of  the  farm  ? 

232.  5th  jr.  2d  T.  Mental.  1.  What  number  must  be  added 
to  38  to  make  54  ?  2.  At  $  1.25  per  yard,  what  will  36  yd. 
of  silk  cost  ?  3.  At  50  ^  a  day  what  will  board  cost  me  for 
the  month  of  July  ?  4.  A  man  had  $  40  and  gave  away 
.12^  of  it.  How  much  did  he  give  away  ?  6.  4  ounces  of 
tea  is  what  decimal  part  of  a  pound  ?  6.  .75  of  a  class  of  44 
were  promoted.  How  many  were  not  promoted  ?  7.  A 
man  bought  a  horse  for  $  96.  He  sold  it  at  a  gain  of  .87^-. 
What  was  the  selling  price  ?  8.  If  f  of  a  doz.  cost  $  1,  how 


§233  SECTION   1.    ELEMENTARY  SCHOOLS  175 

much  will  4  articles  cost  ?  9.  If  f  of  my  money  is  63  ^, 
what  is  ^  of  it  ?  10.  At  2  ^  each  how  many  apples  may  be 
bought  with  2  dimes  and  2  nickels  ? 

Written.  1.  Add  twenty-five  dollars  twenty-five  cents,  three 
hundred  forty-six  dollars  forty-eight  cents,  six  thousand  two 
hundred  forty-two  dollars  seventy-five  cents,  one  thousand 
three  hundred  nine  dollars  twenty-nine  cents,  eighteen  hundred 
dollars  four  cents.  2.  Subtract  28|  from  301|.  3.  The 
multiplicand  is  2-J-,  the  product  is  1^.  What  is  the  multiplier  ? 
4.  I  bought  a  boat  for  $240  and  sold  it  at  a  gain  of  .20. 
Find  the  selling  price.  6.  Divide  3^  by  3^  and  write  the 
answer  in  decimal  form.  6.  If  2^  yds.  of  cloth  cost  16£  ^, 
how  many  yards  can  be  bought  for  63  ^  ?  7.  What  decimal 
of  a  mile  is  1056  ft.  ?  8.  A  man  earns  $  12  a  week ;  how 
much  does  he  earn  a  year  ?  9.  Reduce  to  decimals  and  add, 
i>  3|-£,  f^f,  f$.  10.  Change  to  common  fractions  and  ar- 
range in  order  of  value:  .005,  .62^,  .35,  .33^. 

SIXTH  YEAR 

233.  6th  yr.  ist  T.  Mental.  1.  Change  f  bu.  to  quarts. 
2.  Which  is  greater,  ^  or  -f^  ?  Give  the  difference.  3.  At  4  ^ 
a  pint  what  will  you  pay  for  10  gal.  2  qt.  of  milk  ?  4.  Sold 
for  8  ^  and  lost  2  ^.  What  was  the  per  cent  of  loss  ? 
6.  Have  2  quarters  5  dimes  1  nickel.  How  many  apples  can 
I  buy  at  5  ^  each  ?  6.  A  man  sells  a  carriage  for  $  56,  which 
is  ^  less  than  he  gets  for  his  horse.  What  did  he  receive  for 
the  horse  ?  7.  If  £  of  the  number  of  bushels  in  a  bin  is  20, 
how  many  bushels  in  the  bin  ?  8.  How  many  days  between 
March  12  and  April  15,  1913?  9.  I  had  $3.20  and  spent 
25  %  of  it.  How  much  did  I  spend  ?  10.  An  article  cost 
$  8  and  sold  at  a  gain  of  37|  %.  What  was  the  selling  price  ? 

Written.  1.  Reduce  .027  of  a  ton  to  ounces.  2.  If  a  bin 
contains  5  bu.  3  pk.  2  qt.,  how  much  will  5  bins  of  the  same 
size  contain  ?  3.  A  man  was  born  Dec.  25th,  1852.  How 


176  SECTION   1.    ELEMENTARY  SCHOOLS  §234 

old  was  he  February  12th,  1881  ?  4.  I  had  $  1200.  I  de- 
posited 25  %  of  this  sum  in  the  Bowery  Savings  Bank,  and 
33£  %  of  the  remainder  in  the  Metropolitan  Savings  Bank. 
How  much  cash  have  I  ?  6.  Make  a  bill  for  the  following 
and  receipt  it ;  date  it  to-day ;  make  yourself  the  purchaser. 
Bought  of  A.  J.  Cammeyer :  10  pairs  of  men's  shoes  @  $4.75 ; 
4  pairs  of  boys'  shoes  @  $  1.47.V ;  6  pairs  of  slippers  @  .87^  ^ ; 

9  pairs  of  girls'  shoes  @  $2.43;  8  pairs  of  women's  shoes  @ 
$  3.371.       6.   Paid  $  450  for  a  horse  and  sold  it  at  a  profit  of 
15  %.    What  was  the  selling  price  ? 

234.  6th  yr.  2d  T.  Mental.  1.  It  will  cost  40  ^  to  send  a 
ten-word  telegram  to  Boston.  Every  additional  word  will  cost 
3  t.  What  must  I  pay  for  18  words  ?  2.  If  9  yards  of 
ribbon  cost  $  2.50,  what  will  27  yards  cost  ?  3.  A  man  sold 
25  %  of  his  wheat  to  one  buyer,  12^  %  to  a  second,  and  had 
350  bu.  left.  How  many  bushels  in  the  crop?  4.  A 
dealer  bought  a  second-hand  sofa  for  $40  and  having  spent 

10  %  on  repairs,  sold  it  at  a  gain  of  25  %  on  the  whole  cost. 
For  what  did  he  sell  it  ?       6.   What  is  f  %  of  $  5600  ?     6.    A 
piano  was  marked  $480  but  sold  at  a  discount  of  10  %.     Find 
the  discount.        7.   My  selling  price  for  some  goods  is  $30 
and  my  gain  is  20%.     What  is  my  money  gain?       8.   Lost 
$3    by    selling    a    desk    for    $18.     What    per    cent    lost? 
9.   Find  the  commission  at   5  %  on  400  bbl.  of  fish  at  $  5  a 
barrel.       10.   My  store  is  worth  $  4000.     I  insured  it  for  f  of 
its  value.     Find  the  premium  at  %  %. 

Written.  1.  A  commission  merchant  sold  10  bbl.  of  pears 
for  a  farmer  at  So  a  barrel.  His  commission  was  10  %,  and 
$  2.50  was  charged  for  freight.  What  sum  did  he  return  to 
the  farmer  ?  2.  A  farmer  raised  517  bu.  of  potatoes.  He 
sold  20  °/o  of  them.  How  many  bushels,  pecks,  and  quarts  did 
he  sell  ?  3.  Find  the  value  of  a  rectangular  piece  of  land 
75  rd.  long  and  56  rd.  wide  at  $  144  per  acre.  4.  You  sell  a 
100-lb.  keg  of  horseshoes  for  $14.05  and  gain  25%.  What 


§235  SECTION   1.    ELEMENTARY  SCHOOLS  177 

do  they  cost  you  per  pound  ?  6.  The  bread  from  a  bbl.  of  flour 
(196  lb.)  weighs  31^%  more  than  the  flour.  What  is  the 
weight  of  the  bread  ?  6.  You  pay  $  200  rent  a  month  for 
your  store.  How  many  dollars  of  sales  must  you  make  each 
month  to  raise  this  rent  if  you  average  20  %  profit  ? 
7.  Sold  a  horse  for  $250  and  lost  20%.  What  would  have 
been  the  selling  price  if  I  had  gained  20  %  ?  8.  Jan.  1, 
1913,  John  Smith  bought  of  Henry  Jones  a  piece  of  land 
100  ft.  by  25  ft.  and  paid  for  it  at  50  ^  a  square  foot  by  a 
check  on  the  Tenth  National  Bank,  100  Broadway.  Draw 
the  check. 

SEVENTH  YEAR 

235.  yth  yr.  istT.  Mental.  1.  The  principal  is  $600;  the 
rate,  3%  5  the  time,  2  yr.  6  mo.  What  is  the  interest? 
2.  25  qt.  are  62£  %  of  how  many  gallons?  3.  $17  are  20% 
of  how  many  dollars  ?  4.  A  grocer  received  60  bbl.  of  flour 
and  sold  12  of  them.  What  %  had  he  left  ?  5.  I  collected 
$450  and  received  a  commission  of  4%.  What  was  the  com- 
mission ?  6.  Philadelphia  is  75°  West  longitude.  When  it 
is  noon  in  London,  what  is  the  time  in  Philadelphia  ?  7.  How 
many  dollars  in  £100?  How  many  centimeters  in  3  m  70 
mm  ?  8.  A  piano  listed  at  $400  was  sold  with  discounts  of 
25%  and  10%.  What  was  the  selling  price?  9.  50^  is 
what  per  cent  of  f  of  a  dollar  ?  10.  By  selling  a  book  for 
$4,  I  lost  $1.  Find  the  per  cent  lost. 

Written.  1.  On  May  3d,  1910,  I  borrowed  $640  at  simple 
interest  at  5%.  How  much  do  I  owe  on  Sept.  27th,  1913? 
Count  the  exact  number  of  days.  2.  Change  $  112  to  English 
money.  3.-  What  sum  of  money  invested  at  5%  per  annum 
will  give  an  annual  income  of  $1200?  4.  Two  boys  have 
$48  between  them,  one  having  $18  more  than  the  other. 
How  much  has  each?  6.  A  milkman  sold  86  qt.  of  milk. 
How  many  liters  did  he  sell  ?  6.  Cloth  costing  1200  francs 
in  France  was  shipped  to  America  where  a  duty  of  24%  was 


178  SECTION   1.    ELEMENTARY  SCHOOLS  §236 

levied.  How  much  was  the  total  cost  in  our  money  ?  7.  A 
man  starts  from  Chicago  and  some  days  later  finds  that  his 
watch  is  2  hr.  30  min.  slow.  In  what  direction  has  he  been 
traveling  and  over  how  many  degrees  of  longitude  has  he 
gone  ?  8.  A  recipe  for  1^  Ib.  of  fudge  calls  for  3  cups  of  sugar 
(6  ^  a  pound),  1  tablespoon  of  butter  (32  ^  a  pound),  f  of  a  cup 
of  milk  (9  ^  a  quart),  2  oz.  of  chocolate  (36  X  a  pound),  and  1 
teaspoon  of  vanilla  (10^_1-|-  oz.  bottle).  Find  the  cost  of  a 
pound ;  count  31  tablespoons  to  a  pint  or  pound,  2  cups  to  a 
pint,  and  3  teaspoons  to  a  tablespoon. 

236.  7th  yr.  2d  T.  Mental.  1.  Principal  is  what  when  rate 
is  5%  ;  time  is  2  yr.  6  mo.;  interest  is  $12.50?  2.  Principal 
is  $  600 ;  time  is  3  yr.  4  mo. ;  interest  is  $  80.  What  is  the 
rate?  3.  Principal  is  $ 900 ;  rate  5%;  interest  $60.  What 
is  the  time  ?  4.  The  simple  interest  on  a  certain  principal 
is  $  146.  What  is  the  exact  interest  ?  5.  If  11  books  cost 
$  1.32,  what  will  100  books  cost  ?  6.  9,  + 12,  -=-  3,  x  12, 
-  3,  --  9,  +  4,  -  11  ?  7.  x,  x  12,  +9,  -=-3,  +9,  - 12,  the 
result  is  20,  find  x.  8.  The  following  recipe  is  for  a  dozen 
biscuits.  How  much  of  each  ingredient  must  be  used  to  make 
15  biscuits  ?  For  a  dozen,  2  cups  of  flour,  %  cup  of  milk,  ^ 
teaspoon  salt,  21  teaspoons  of  baking  powder,  2  tablespoons  of 
shortening.  9.  Find  the  area  of  a  triangle  whose  base  is  12 
in.  and  whose  altitude  is  10  in.  10.  What  is  the  ad  valorem 
duty  on  100  yd.  of  silk  valued  at  $4  a  yard,  duty  25%? 

Written.  1.  A  merchant  had  $11,640;  he  invested  26|% 
of  it  in  dry  goods  and  12 £%  of  the  remainder  in  groceries. 
How  much  money  had  he  left  ?  2.  Find  the  interest  of 
$320  for  1  yr.  8  mo.  27  days  at  5%.  3.  Find  the  amount  of 
$ 750  from  Jan.  12th,  1903,  to  Dec.  16th,  1905,  at  1%.  Count 
the  exact  number  of  days.  4.  Multiply  six  hundred  twenty- 
five  thousandths  by  twenty-five  millionths  and  divide  the 
product  by  one  hundred  twenty-five  hundred  thousandths. 
6.  John  Smith  bought  $400  worth  of  goods  from  Thomas 


§237  SECTION   1.    ELEMENTARY  SCHOOLS  179 

Brown  on  3-raonths'  credit.  Write  a  bank  note  for  the  trans- 
action, dating  it  to-day.  6.  Find  the  proceeds  at  bank  dis- 
count @  6%  if  the  note  is  discounted  15  days  after  to-day. 
7.  If  29  bales  of  hay  are  consumed  by  58  cows,  how  many 
bales  will  last  46  cows  for  the  same  time  ?  8.  Sold  a  wagon 
for  $420,  which  is  16%  less  than  the  cost.  What  should  I 
have  sold  it  for  to  gain  25%  ?  9.  What  number  increased 
by  |  of  itself  equals  402  ?  10.  Find  a  :  3x  —  5  (a? +  3)  =  —  21. 

EIGHTH  TEAR 

237.  8th  Yr.  ist  T.  Mental.  1.  Reduce  3  dm  7m  5  cm 
to  mm.  2.  A  house  is  valued  at  $6000.  It  is  insured  for 
76  %  of  the  valuation  at  20  /  per  S 100.  What  is  the  premium  ? 
3.  15,  +  17,  H-  4,  square,  -  1,  X  11  =  ?  4.  Eeduce  2  T.  5 
cwt.  to  pounds.  6.  Sold  a  wagon  for  $72,  gaining  12i%. 
What  was  the  cost  ?  6.  How  many  pint  bottles  can  be  filled 
by  3  gal.  3  qt.  of  wine  ?  7.  What  is  the  perimeter  of  a  square 
whose  area  is  121  sq.  yd.  ?  8.  What  is  the  circumference  of 
a  28-inch  wheel  ?  9.  What  is  the  ratio  of  4  ft.  2  in.  to  5  ft.  ? 
10.  State  which  of  the  following  are  multiples,  which  factors, 
which  powers,  and  which  roots  of  8 :  16,  4,  64,  24,  2. 

Written.  1.  If  .625  of  a  cord  of  wood  cost  $3.75,  what  will 
.75  of  a  cord  cost  ?  2.  How  much  will  it  cost  to  carpet  a  room 
14  ft.  by  12  ft.  with  carpet  f  of  a  yard  wide  @  $  1.25  a  yard? 
Strips  to  run  lengthwise.  3.  I  gained  1  °/0  by  selling  my  farm 
for  $1346.70.  What  was  the  cost?  4.  A  railroad  passes 
through  a  farm  taking  a  strip  \\  miles  long  and  66  ft.  wide. 
What  is  the  value  of  this  land  at  $  80  an  acre  ?  5.  A  60-day 
note  of  $  600,  dated  Aug.  4th,  1913,  was  discounted  Sept.  1st 
@6%.  Find  the  proceeds.  6.  If  24  men  in  12  days  of  9 
hours  each  can  do  a  certain  amount  of  work,  how  many  days 
of  8  hours  each  will  it  take  36  men  to  do  twice  that  amount  of 
work  ?  7.  A  tank  full  of  water  is  8  ft.  x  4  ft.  x  3  ft.  What 

x  —  1 
is  the  weight  of  the  water  ?       8.   Find  x:  2x —  =  30. 


180  SECTION   1.    ELEMENTARY  SCHOOLS 

238.    8th  Yr.  2d  T.     Mental     1.   35,  +  40,  -=-  5,  x  20,  -h  4, 

-4-  10,  —  5  =  ?  2.  How  many  bottles  f  qt.  in  capacity  can 
be  filled  from  a  demijohn  containing  4|  qt.?  3.  How  many  lb. 
in  1000  Kg.?  How  many  tons?  4.  Sold  a  book  for  60^, 
gaining  20%.  What  %  would  I  have  lost  had  I  sold  it  for 
40^  ?  6.  The  cost  of  a  bicycle  is  $  36.  What  shall  the  marked 
price  be  to  allow  a  gain  of  16f  %,  after  falling  33^%  from  the 
marked  price  ?  6.  How  many  miles  in  80  Km  ?  7.  Solve 
8  x  —  3  =  5  x  +  21.  8.  Sold  a  horse  for  $  225  and  gained  $  25. 
What  per  cent  gained  ?  9.  £  +  £  of  my  money  equals  $  140. 
How  much  money  have  I  ?  10.  If  6  men  can  do  a  piece  of 
work  in  18  days,  how  long  will  it  take  4  men  to  do  the  same 
work  ? 

Written.  1.  Add  491673,  28674,  847598,  892654,  34567, 
67891,  391638,  328695,  64738  and  473876.  2.  Solve  by  short 
processes :  (a)  Find  the  cost  of  5400  lb.  @  $  6.50  per  ton ; 
(6)  What  is  250  %  of  400  ?  3.  The  surface  of  a  sphere  is 
1963.5  sq.  in.  What  is  its  diameter?  4.  Find  the  cost  of 
carpeting  a  room  12  ft.  by  9  ft.  with  carpet  f  yd.  wide  at  95^ 
a  yd.  The  strips  are  to  run  lengthwise.  5.  A  90-day  note  for 
$  1000  with  interest  at  7  %  was  dated  Jan.  17th,  1907,  and  dis- 
counted March  2d,  at  6%.  Find  the  proceeds.  6.  The  per- 
imeter of  a  square  piece  of  land  is  16  rods.  How  much  is  it 
worth  at  10  cents  a  square  foot?  7.  A  cylindrical  cistern 
with  a  diameter  of  5  ft.  has  27  inches  of  water  in  it.  How 
many  gallons  are  there  ?  8.  When  it  is  7  A.M.  in  New  York 
City,  what  time  is  it  in  London  ?  9.  I  bought  120  meters  of 
lace  at  4  francs  per  meter.  For  what  must  I  sell  it  per  yard, 
U.  S,  money,  in  order  to  gain  20  %  on  the  investment  ?  10.  I 
have  $  6600  with  which  to  make  an  investment.  I  am  offered 
6%  stock  @20%  premium  or  5%  stock  at  10%  premium. 
Which  is  better  and  by  how  much  annually  ? 


SECTION   2.     PRIMARY  LICENSE— CITY 

239.  Scope.     This  section  gives  an  idea  of  the  prepara- 
tion necessary  for  obtaining  a  certificate  to  teach  arith- 
metic in  elementary  schools.     With  a  few  exceptions  the 
following  tests  have  been  given  for  License  No.  1,  a  license 
to  teach  the  requirements  of  the  first  six  years  in  the  school 
system  of  New  York  City. 

240.  Algebra.     1.   If  an  automobile  was  sold  for  $  1025  at 
a  profit  of  25  % ,  how  much  did  it  cost  ?     Use  x  or  other  alge- 
braic symbol. 

2.  Illustrate  the  correct  use  of  the  equation  in  solving  two 
easy  examples  in  percentage. 

3.  State  and  solve  an  easy  problem  in  simple  interest  to  find 
the  rate.     Use  x. 

241.  Apperception.     4.  Apply  the  principle  of  apperception 
to  a  first  lesson  in  decimals. 

NOTE.  This  calls  for  a  knowledge  of  the  five  formal  steps  of  the 
Herbartians :  preparation,  presentation,  comparison,  generalization, 
and  application.  These  steps  are  taken  in  §  149.  The  reader  should 
find  them. 

242.  Cancellation.     6.   State  the  principle  of  cancellation 
and  show  how  it  should  be  taught. 

6.  After  stating  two  principles  upon  which  the  process  of 
cancellation  is  based  in  the  multiplication  of  fractions,  show 
how  that  process  should  be  taught. 

181 


182  SECTION  2.    PRIMARY  LICENSE  — CITY  §243 

243.  Decimals.     7.   State  the  steps  in  teaching  the  rule  for 
the  multiplication  of  a  decimal  by  a  decimal. 

8.    State  the  steps  in  teaching  the  rule  for  the  division  of  a 
decimal  by  a  decimal. 

244.  Denominate  Numbers.     9.   With  respect  to  a  single 
exercise  in  linear  measurement  suitable  for  the  second  year, 
state :  (a)  what  is  to  be  measured ;  (6)  what  it  is  to  be  meas- 
ured with ;  (c)  how. 

10.   How  should  the  subject  of  cubic  measure  be  presented  ? 

245.  Developments.     11.    Enumerate  the  45  combinations 
of  digits  (taken  two  at  a  time  in  addition)  which   must   be 
learned  in  the  first  two  years  of  school  and  state  how  you 
would  teach  these  combinations. 

12.  Explain  "Only  like  numbers  can  be  subtracted";  show 
how  this  principle  applies  in  the  subtraction  of  integers,  of 
common  fractions,  of  decimals,  and  of  denominate  numbers. 

13.  (a)  Explain  the  Austrian  method  of  subtraction,  using 
the  following  example:  subtract  58  from  100.     (6)  State  its 
advantages  and  its  disadvantages. 

14.  Show  the  complete  form  of  blackboard  demonstration 
of  the  process:  (a)  of    adding   three-figure   numbers;    (&)  of 
multiplying  by  three-figure   numbers  ;    (c)  of   dividing  by  a 
one-figure  number. 

246.  Devices.     15.   Suggest  a  device  for  helping  a  pupil  to 
remember  what  17  minus  8  is. 

16.    Suppose  a  pupil  forgets  the  product  of  6  and  7;  suggest 
three  devices  which  may  be  helpful  to  him. 

247.  Diagrams.     17.    Show  by  a  diagram  that   multiplier 
and  multiplicand   (when   neither  is  concrete)   can  be   inter- 
changed without  altering  the  product. 

NOTE.     This  is  the  commutative  law  in  multiplication. 


§248  SECTION  2.    PRIMARY  LICENSE  — CITY  183 

18.  Solve  by  use  of  a  diagram :  How  many  yards  of  cloth 
at  S  f  a  yard  can  be  bought  for  $  6  ? 

19.  Show  graphically  that  f  (of  1)  is  equal  to  3  -5-  5. 

20.  State  a  problem  in  profit  and  loss,  given  the  gain  and 
gain  per  cent,  and  requiring  the  selling  price :  (a)  Solve  the 
problem  stated.     (6)  Show  how  to  illustrate  the  solution  by  a 
drawing. 

248.  Drills.      21.    "Since   memory   is  served   by   multiple 
associations  quite  as  well  as  by  repetition,  the  drills  employed 
should  be  varied  in  form,  iii  content,  and  in  mode  of  applica- 
tion."    In  the  light  of  the  foregoing  quotation,  suggest  three 
distinct  types   of  drill  under  each  of   the   following   heads: 
(a)  counting;   (6)  multiplication. 

22.  Describe  three  devices  or  modes  of  procedure,  for 
enabling  the  teacher  to  conduct  efficiently  a  drill  in  rapid 
addition,  and  state  the  advantages  of  each. 

249.  Errors.      23.    Following,   there   are  common   errors. 
Correct  each  and  explain  its  nature  as  to  a  pupil :    (a)  The 
area  of  a  rectangle  is  12  inches  multiplied  by  6  inches  or  72 
square  inches.     (6)  Since  39  is  1  less  than  40,  it  may  be  writ- 
ten IXL  by  the  Roman  notation,     (c)  3  is  contained  in  15  ^ 
5ft  times,     (d)  5±  gallons  may  be  taken  from  45  gallons  8 
times    and  ^-  gallons   will    remain,  because   45  -+-  5£  =  8^-. 
(e)  6%=  144,  1%  =  24,  100%  =  2400. 

250.  Fractions.      24.   Discover  the  effect  on  a  fraction  of 
dividing  both  numerator  and  denominator  by  the  same  number. 

25.  Discover  the  rule  for  multiplying  a  fraction  by  a  fraction. 

26.  Discover  the  rule  for  dividing  a  fraction  by  a  fraction. 

27.  Tell  in  order  the  steps  involved  in  getting  the  pupils  to 
understand  the  reduction  of  a  common  fraction  to  a  decimal. 
Give  illustrations. 


184  SECTION  2.    PRIMARY  LICENSE  — CITY  §251 

28.  State  and  illustrate  the  steps  in.  simplifying  a  complex 
fraction.     Define  a  complex  fraction. 

29.  Find  the  answer  to  the   following  problem :  ^  of  f  is 
75  v/o  of  what  number  ?     Show  how  you  would  have  children 
understand  each  step  of  the  process. 

30.  State  a  complete  problem  which  involves  finding  what 
part  one  common  fraction  is  of  another.     State  its  solution  by 
a  diagram. 

31.  To  find  a  number  when  the  number  plus  or  minus  a  part 
of  itself  is  given.     State   a  practical   problem  of   each  type 
designated. 

251.  Games.     32.    Describe  two  games  or  two  other  recrea- 
tive'exercises  appropriate  to  the  first  or  second  year  of  school 
and  involving  counting  or  other  number  work. 

252.  Geometry.      33.   How  would  you  teach  finding  the 
area :  (a)  of  a  rectangle  ?    (6)  of  a  parallelogram  ?    (c)  of  an 
oblique  triangle  ?     (d)  of  a  trapezoid  ? 

34.  How  would  you  teach  finding  the  circumference  of  a 
circle  ? 

253.  Helps.     35.    A  pupil  cannot  solve  this  problem:  Iff 
of  a  yard  costs  24  ^,  how  much  will  1  yard  cost  ?     Help  him. 

36.  At  3  ^  each  how  many  apples  can  be  bought  for  12^? 
A  pupil  does  not  understand  why  the  number  of  apples  for 
12^  is  the  number  of  times  3^  is  contained  in  12^.  Help 
him. 

254.  Inductive   Exercises.     37.   What  is  the  gist  of  the 
inductive  method  ? 

38.  Show  inductively  (as  in  second  year)  that -adding  10  to 
both  subtrahend  and  minuend  does  not  change  the  difference. 

39.  Discover   inductively  a  rule  for   the   divisibility  of  a 
number  by  9. 


§255  SECTION  2.    PRIMARY  LICENSE  — CITY  185 

255.  Interest.     40.    State  and  solve  a  practical  problem  in 
simple  interest  to  find  the  time. 

41.  (a)  Compose  a  problem  in  bank  discount  involving  the 
discount  of  a  non-interest  bearing  note.  (6)  Write  the  note, 
(c)  Give  the  steps  in  its  solution. 

256.  L.  C.  D.     42.    Add  |,  f,  f,  |.     (a)   Find  by  inspection 
the  least  common  denominater.     (6)  Give  the  rule  for  finding 
the  least  common  denominator  by  inspection. 

43.  Explain  clearly  how  to  find  the  least  common  denom- 
inator when  the  denominators  are  too  large  for  the  method  by 
inspection,  as  in  the  case  of  ^,  -fa,  -^j. 

257.  Logical   Division.      44.    Discriminate  and  illustrate 
logical  division  ;  logical  definition. 

45.  In  beginning  the  subject  of  long  division,  if  the  follow- 
ing divisors  are  to  be  used,  in  what  order  should  they  be  used 
and  why  ?     24,  17,  29,  27,  80  ? 

46.  In  presenting  the  subtraction  of  one  mixed  number  from 
another,  what  is  the  simplest  type  of  this  case  ?     Illustrate 
this  type  and  three  others  of  increasing  difficulty  suitable  for 
early  lessons  in  the  subject. 

47.  State  in  the  order  in  which  they  should  be  taught  the 
types  of  examples  in  division  which  involve  one  or  more  deci- 
mal numbers.     Give  reasons  for  the  order  chosen. 

258.  Longitude  and  Time.     48.    In  a  first  lesson  on  longi- 
tude what  points  should  be  brought  out?     What  device  should 
be  employed? 

49.  How  would  yo\i  teach  that  the  time  of  a  place  farther 
west  is  earlier? 

60.  What  is  the  difference  in  time  between  a  place  whose 
longitude  is  54°  east  and  a  place  whose  longitude  is  60° 
west  ? 

NOTE.     The  device  required  is  to  locate  the      w— 
places  as  suggested  at  the  right. 


186  SECTION   2.    PRIMARY  LICENSE  — CITY  §259 

259.  Percentage.     61.    Taking  some  one  activity,  industry, 
or  experience  as  a  center,  construct  about  it  five  practical  prob- 
lems involving  different  applications  of  percentage. 

62.  State  and  solve  a  problem  in  trade  discount. 

63.  A  man  bought  5  %  stock  and  thereby  secured  an  income 
of  6  °/o  on  his  investment,     (a)  -How  much  did  he  pay  for 
the  stock?      (6)   Explain   as   to  a   pupil   how   to  prove  the 
answer. 

260.  Proofs.     64.   What  is  meant  by  proving  an  example  ? 
by  proving  a  problem  ? 

65.  Give  reasons  why  pupils  should  be  taught  checking  of 
results  in  the  four  operations  and  tell  what  these  methods  are. 

66.  In  the  case  of  the  following  processes,  state  and  exem- 
plify modes  of  verifying  or  checking  results  which  are  suitable 
to  pupils  below  the  seventh  year :  (a)  addition  (two  modes) ; 
(6)  finding  a  whole  when  a  fractional  part  is  given ;  (c)  reduc- 
tion ascending. 

67.  Solve  and  prove  an  original  example  in  the  second  or 
third  case  of  percentage. 

261.  Proportion.     68.    State  and  solve  a  practical  problem 
in  proportion. 

59.  (a)  Give  two  examples  involving  ideas  of  ratio  or  of 
proportion,  appropriate  for  the  third  or  fourth  year.  Give 
model  analysis.  (6)  Write  a  practical  problem  in  direct  pro- 
portion and  one  in  inverse  proportion,  (c)  Define  ratio ;  define 
proportion. 

262.  Unit  of  Measure.     60.   (a)  What  is  meant  by  unit  oi 
measure?     (6)  State  and  solve  a  problem  in  which  the  num- 
ber 3  may  be  used  as  a  unit. 

263.  Use  of  Text-book.     61.   State  how  the  text-book  should 
be  used  in  arithmetic  —  as  in  the  case  of  fractions. 


SECTION  3.     PRIMARY  LICENSE  —  STATE 

264.  Scope.     The  scope  of  this  section  is  about  the  same 
as  that  of  the  last.     These  tests  have  been  given  by  the 
Department  of  Education  of  the  State  of  New  York,  for 
Training  School  Certificates  or  certificates  to  teach  in  any 
elementary  school  of  the  state  except   in  certain  of   the 
large  cities. 

265.  Aliquot  Parts.     1.  Give  the  aliquot  parts  that  in  your 
judgment  should  be  memorized  during  the  elementary  course. 
When  should  they  be  learned  ?     Make  up  two  examples  of 
different   types,  involving  in  their    solution  a  knowledge  of 
aliquot  parts;  assume  that  the  examples  are  to  be  solved  men- 
tally by  pupils  of  the  eighth  grade. 

XOTE.  An  aliquot  part  is  understood  to  mean  one  of  the  equal 
parts  of  100 ;  the  word  means  some  times.  Thus,  the  aliquot  parts 
(of  100)  are  50  (one  of  two  parts),  33£  (one  of  three),  25  (one  of 
four),  20  (one  of  five),  16|  (one  of  six),  and  so  on. 

266.  Analysis.     2.   Of  what  value  are  forms  of  analysis  ? 
In  what  respect  are  forms  of  analysis  serviceable  to  you  per- 
sonally in  the  solution  of  problems? 

3.  Give  a  model  analysis:    A  bookseller   sold   a  book   for 
$  L'.25  at  a  loss  of  10  per  cent;  how  much  did  he  lose  ? 

4.  A  man  sold  4  of  his  farm  for  what  the  whole  of  it  cost ; 
what  percent  did  he  gain  on  the  part  sold?     Give  a  model 
analysis. 

5.  Give  a  logical  analysis  :  A  man  bought  stock  paying  4  % 
dividends,  at   20%  discount;    what   rate   of   income   did   he 
receive  on  the  investment? 

187 


188  SECTION  3.    PRIMARY  LICENSE  — STATE  §267 

6.  What  must  be  the  marked  price  of  a  hat  costing  $  6,  so 
that  after  discounting  the  price  30  °Jo  the  dealer  may  make  a 
profit  of  16|  %  ?  Analyze. 

267.  Developments.     7.   Outline  a  development  lesson  to 
teach  the  terms  multiplicand,  multiplier,  and  product. 

8.  Show  how  to  present  inductively  the  idea  of  a  common 
fraction. 

NOTE.  '  Inductively '  is  to  be  interpreted  as  '  objectively.'  Strictly 
speaking  only  principles  or  laws  can  be  presented  inductively. 

9.  When  should  decimal  fractions  be  first  introduced  and 
how  should  they  be  presented  ? 

10.  Explain  how  the  difference  between  a  draft  and  a  eheck 
may  be  made  clear  to  a  class. 

11.  Explain  how  the  difference  between  bonds  and  stocks 
may  be  made  clear  to  a  class. 

268.  Proofs.     12.   Write  check  problems  to  be  used  to  test 
the  correctness  of  the  solution  of  the   following   examples : 
(a)  If  a  train  runs  32  miles  in  one  hour,  how  far  will  it  run  in 
45  minutes  ?     (&)  What  is  the  interest  on  $  425  for  2  years 
and  4  months  at  6  %  ? 

269.  Teaching.      13.   Write  three  examples  such  as  you 
would  use  in  a  first  lesson  on  long  division  and  show  how  you 
would  teach  the  subject. 

14.  Show  by  an  outline  how  you  would  present  addition  of 
fractions. 

16.  Show  how  you  would  teach  by  means  of  a  rectangle  or 
a  circle  that  \  x  \  =  y^.  Suggest  what  more  you  would  do  to 
develop  the  rule  for  the  multiplication  of  fractions. 

NOTE.  The  complete  method  (§  20)  is  required.  The  one-line 
diagram  (§  12)  is  superior  to  a  rectangle  or  a  circle ;  fractions  with 
numerators  other  than  1  are  to  be  preferred. 


§270  SECTION  3.    PRIMARY  LICENSE— STATE  189' 

16.  Illustrate  how  you  would  explain  to  a  class  the  reason 
for  inverting  the  divisor  in  the  -division  of  fractions. 

17.  State  four  cases  of  problems  in  simple  interest  and  give 
a  plan  for  teaching  one  of  them. 

270.  Theory.  18.  How  and  why  should  the  aim  of  arith- 
metic teaching  in  the  primary  grades  differ  from  the  aim  in 
grammar  grades  ? 

19.  Discuss  the  extent  to  which  problems  calling  for  the 
application  of   principles  should   be  given  in  the  first   three 
grades. 

20.  Explain  the  principal  advantages  claimed  for :  (a)  the 
Austrian    method   of    teaching    subtraction ;    (&)    the    spiral 
method  of  teaching  fractions  and  other  topics  in  arithmetic. 

NOTE.  The  spiral  plan  calls  for  teaching  the  leading  principles 
with  fractions  which  have  small  denominators,  and  then  in  teaching 
the  whole  subject  again  with  fractions  which  have  larger  denomina- 
tors, and  so  on,  making  each  spiral  more  comprehensive  than  the  one 
before. 

21.  State  arguments  against  the  use  of  the  Grube  method. 

NOTE.  The  Grube  method  teaches  the  fundamental  operations 
with  each  number  before  taking  up  the  next  higher.  Thus,  the  fol- 
lowing exercises  with  4  are  taught  before  5:  2  +  2  =  4,  3  +  1  =  4, 
1  +  1  + 1  +  1  =  4;  4-1  =  3,  4-3  =  1;  2x2=4,  4x1  =  4,  1x4  =  4; 
4 '-5- 1  =  4,  4  -s-  4  =  1,  4  -  2  =  2,  J  of  4  =  1,  \  of  4  =  2. 

22.  Discuss  the  merits  of  the  solution  of  the  examples  in 
simple  interest  by  formula,    as   compared  with  the  value  of 
solving  them  by  the  application  of  the  proper  process  of  rea- 
soning without  the  use  of  formula. 


SECTION  4.     HIGHER  LICENSES 

271.  Scope.     This  section  gives  an  idea  of  the  prepara- 
tion necessary  to  secure  a  higher  license  to  teach  in  the 
schools  of  New  York  City.     These  tests  have  been  given 
for  promotion  license  (license  for  the  7th  and  8th  years), 
license  for  head  of  department,  license  for  assistant  to 
principal,  or  license  for  principal. 

272.  Fundamentals.     1.  Discuss  these  definitions  :  A  num- 
ber is  a  unit  or  collection  of  units.     A  number  is  an  abstract 
ratio  of  one  quantity  to  another  of  the  same  kind. 

NOTE.  How  many  is  measured  by  comparison.  There  is  a  coin 
for  this  eye  and  a  coin  for  this  eye ;  there  are  as  many  coins  as  a  man 
has  eyes ;  '  as  many  as  a  man  has  eyes '  is  named  two.  Number  is  an 
expression  of  how  many  (§  77). 

There  is  an  act  of  comparison  for  each  individual  or  there  are  as 
many  times  of  comparison  as  individuals.  Thus  there  are  '  2  coins 
1  time '  or  '  1  coin  2  times.'  The  number  of  times  of  comparison  is 
called  ratio.  A  number  is  a  ratio. 

An  '  act  of  comparison '  or  a  '  time '  is  an  individual.  Hence,  the 
definition,  '  Number  is  an  expression  of  how  many '  includes  the  defi- 
nition, 'Number  is  a  ratio';  the  first  refers  to  individuals  of  any 
kind  while  the  second  refers  to  individuals  alone  that  are  acts  of 
comparison. 

2.   Describe  and  criticise  the  Speer  method. 

NOTE.  The  Speer  method  is  based  upon  the  ratio  idea  of  number. 
Thus,  the  teacher  shows  two  lines,  two  surfaces,  or  two  solids,  one  of 
which  (.4)  contains  the  other  (B)  3  times.  A  is  how  many  times  Bl 
3  times.  What  is  the  ratio  of  A  to  B  ?  3.  If  B  is  the  cost  of  1  apple, 
what  is  A  1  The  cost  of  3  apples.  B  is  what  part  of  A  ?  ^.  What 

190 


§  272  SECTION  4.    HIGHER  LICENSES  191 

is  the  ratio  of  B  to  A  ?  \.  If  A  is  the  cost  of  3  apples,  what  is  5? 
The  cost  of  1  apple.  What  is  the  cost  of  3  apples  at  5^  each  ?  The 
ratio  of  3  apples  to  T  apple  is  3 ;  the  sum  which  bears  to  5  ^  the  ratio 
of  3  is  15  ^  ;  the  cost  of  3  apples  is  15  f. 

3.  Describe  and  criticise  the  McLellan  and  Dewey  method. 
NOTE.     This  method  is  based  upon  the  ratio  idea  of  number.     A 

ratio  involves  the  quantity  to  be  measured,  the  unit  of  measure, 
and  the  times  contained  (number).  Thus,  in  6t :  2  $  =  3,  6?  is  the 
quantity  to  be  measured,  2  ^  is  the  unit  of  measure,  and  3  is  the  num- 
'ber.  What  is  the  cost  of  3  apples  at  5?  each?  Since  the  unit  of 
measure  is  5  ^  and  the  number  is  3,  the  quantity  to  be  measured  is 
3  times  5^  or  15^. 

4.  "  The  ratio  idea  of  number  should  be  introduced  early, 
and   applied  to  the   work   with   fractions."  —  D.   E.   Smith, 
(a)  What  is  meant  by  the  '  ratio  idea  of  number '  ?      (6)  By 
giving  two  typical  problems,  illustrate  the  use  of  the  ratio  idea 
in  the  early  teaching  of  arithmetic,    (c)  By  giving  three  typical 
problems,  illustrate  its  use  in  fractions. 

6.  Give  two  meanings  of  the  expression  f .  Show  graphically 
their  equivalency. 

6.  What  is  meant  by  unitary  analysis  in  arithmetic  ?   Illus- 
trate by  a  problem. 

NOTE.  At  3  for  5  ^,  what  is  the  cost  of  5  apples  ?  Unitary  analysis 
requires  the  finding'of  the  cost  of  1  apple. 

7.  In  arithmetic,  give  and  illustrate  the  laws  of  association, 
commutation  and  distribution. 

NOTE.  The  commutative  (interchangeable)  law  applies  to  addi- 
tion and  to  multiplication.  Addends  may  be  interchanged  without 
affecting  their  sum ;  factors,  without  affecting  their  product.  Thus, 
6  +  8  =  8  +  6;  6x8  =  8x6. 

The  associative  (bringing  together)  law  applies  to  addition  and 
to  multiplication.  Addends  may  be  grouped  in  any  way  without 
affecting  their  sum ;  factors,  without  affecting  their  product.  Thus, 
2  +  (3  +  4)  =  3  +  (2  +  4)  =  4  +  (2  +  3) ;  2  x  (3  x  4)  =  3  x  (2  x4) 
=  4  x  (2  x  3). 


SECTION  4.    HIGHER  LICENSES  §  272 

The  distributive  (taking  apart)  law  applies  to  addition  and  multi- 
plication combined.  A  product  is  the  sum  of  the  products  of  the 
parts  of  the  multiplicand  by  the  multiplier ;  a  product  is  the  sum  of 
the  products  of  the  multiplicand  by  the  parts  of  the  multiplier.  Thus, 
287  x  3  =  200  x  3  +  80  x  3  +  7  x  3 ;  287  x  23  =  287  x  3  +  287  x  20. 

8.  Describe  how  the  following  units  are  derived  or  fixed : 
meter ;  are ;  liter  ;  grain ;  yard ;  gallon  ;  pound ;  dollar ;  leap 
year. 

9.  Give  the  unit  of  the  metric  system  which  most  nearly 
corresponds  to  the  following :  inch;  ton;  mile;  gallon;  grain. 
Give  the  equivalent  of  each. 

10.  State  the  arguments  in  favor  of  beginning  number  work 
with  counting,  and  against  the  system  of  beginning  with  num- 
ber pictures. 

11.  Show  briefly  how  the  simple  equation  may  be  made  part 
of  elementary  arithmetic,  indicating  the  topics  to  which  it  is 
applicable. 

12.  Give  reasons  for  or  against  the  use  of  cases,  rules,  and 
formulas  in  teaching  percentage. 

13.  Give  the  rule  or  formula  and  explain  an  objective  illus- 
tration that  would  make  it  clear  to  pupils  :  (a)  for  the  area  of 
a  circle;  (6)  for  the  volume  of  a  sphere. 

14.  Concerning  the  calculating  of  interest  name  one  typical 
legal  enactment  or  business  custom  which  requires  that  time 
and  time  rate  be  estimated:  (a)  by  compound  subtraction  and 
using  360  da.  to  a  year ;  (b)  by  finding  the  exact  number  of 
days  and  counting  360  da.  to  a  year ;  (c)  by  finding  the  exact 
number  of  days  and  counting  365  da.  to  a  year. 

NOTE,  (a)  U.  S.  rule  for  partial  payments;  (ft)  bank  discount; 
(c)  transaction  with  U.  S.  government* 

15.  Illustrate  short  process :  (a)  to  multiply  by  25 ;    (6)  to 
multiply  by  39 ;  (c)  to  divide  by  125 ;  (d)  to  divide  by  16|; 
(e)  to  add  fractional  units. 


§272  SECTION  4.    HIGHER  LICENSES  193 

16.  (a)  State  in  the  form  of  a  syllogism  the  argument  in- 
volved in  the  explanation:     If  8  gal.  cost  $2.40,  how  many 
gallons  can  be  bought  for  $  5.40  ?    (6)  State  two  ways  in  which 
the  argument  may  be  shortened  by  the  omission  of  a  premise. 

NOTE.  The  no.  gallons  is  8  gal.  multiplied  by  the  no.  of  times 
$2.40  is  contained  in  $5.40.  The  no.  of  times  $2.40  is  contained 
in  &  5.40  is,  etc. 

17.  State  the  axioms  or  general  laws  of  number  on  which 
rests  the  process  for  finding  G.  C.  D.  of  two  numbers  by  the 
division   of  the  greater  by  the   less,  the   divisor  by  the  re- 
mainder, etc.  (Euclidean  method). 

NOTE.  A  factor  of  each  of  two  numbers  is  a  factor  of  their  sum, 
or  of  their  difference.  A  factor  of  a  number  is  a  factor  of  any  multi- 
ple of  that  number. 

18.  Show  how  the  process  of  multiplying  numbers,  involv- 
ing decimals,  may  be  explained  through  the  fundamental  prin- 
ciples of  the  decimal  notation  without  referring  to  common 
fractions. 

19.  Name  five  topics  in  arithmetic  that  can  be'taught  with- 
out giving  all  the  reasons,  and  explain  in  each  case  what  device 
you  would  use  to  justify  your  action  to  your  class. 

20.  Deduce  the  formula  for  changing  given  temperature  m° 
from  F  to  (7;  R  to  F.      The  freezing  point  and  the  boiling 
point  are  as  follows  :  F,  32°-212°  ;  C,  0°-100°  ;  R,  0°-80°. 

21.  What  is  the  test  of  divisibility  of  a  number  by  3  ?     By 
4  ?     By  6  ?     By  45  ? 

22.  Show  how  to  find  the  G.  C.  D.  and  L.  C.  M.  of  several 
fractions,  as  f,  f,»f,  etc. 

NOTE.  The  G.  C.  D.  of  8  12&s,  9  12ths  and  10  12ths  is  1  12th ;  the 
L.  C.  M.  is  360  12ths  or  30. 

23.  What  is  meant  by  rate  of  exchange  ?     Explain  why  it 
varies. 


194  SECTION  4.    HIGHER  LICENSES  §273 

24.  Describe  a  good  method  of  conducting  a  recitation  in : 
(a)  written  arithmetic  ;  (&)  oral  arithmetic. 

25.  Prove  that  7,  11, 13  are  factors  of  all  numbers  composed 
solely  of  the  first  and  fourth  orders  (of  the  decimal  system) 
taken  in  equal  amounts  as  6006,  8008. 

273.  Problems  and  Examples.  26.  How  much  water 
must  be  added  to  a  5  %  solution  of  medicine  to  get  a  1  °/0  so- 
lution ? 

27.  If  the  list  price  is  60  %  more  than  the  cost,  what  addi- 
tional %  of  discount,  besides  the  customary  discount  of  25  % 
to  the  trade,  may  be  allowed  for  cash  payment  in  order  to  gain 
14  %  by  the  sale  ? 

NOTE.  L,  f  C  to  manufacturer ;  1st  rem,  £  L ;  2d  dis,  x  % ;  2d  rem 
or  SP,  -  —  x  f  L ;  SP,  114  %  C  to  manufacturer;  x  =  ? 

28.  For  what  sum  must  I  draw  my  note  of  3  mo.  to  yield 
$  1000  at  5  %  bank  discount  ? 

29.  "Find  the  cost  of  100  Km  of  wire,  55  cm  in  diameter,  at 
1  franc  24  centimes,  per  Kg,  the  specific  gravity  of  the  wire 
being  8.8. 

30.  (a)  Find  the  contents  of  a  cylindrical  tank  8  m  long 
and  9  dm  in   diameter;     (6)  express  its  contents  in  liters; 
(c)   express  in  kilograms  the  weight  of  water  at  maximum 
density  which  this  tank  will  hold;  (d)  express  in  kilograms 
the  weight  of  the  same  quantity  of   oil,  specific  gravity  .7; 
(e)  translate  the  answers  to  (a),  (&),  and  (c)  into  their  approxi- 
mate English  equivalents  (cubic  inches,  quarts,  pounds). 

31.  New  York  is  longitude  73°  58'  25.5"  ^W.,  Sydney,  Aus- 
tralia, is  151°  12'  39"  E.     When  it  is  9  A.M.,*March  16,  at  New 
York,  what  is  the  time  at  Sydney  ? 

32.  A  certain  calculating  machine  can  give  prodxicts  up  to 
ten  digits.     Explain  how  to  use  it  to  multiply  73,924,583  by 
762,343. 


§  273  SECTION  4.    HIGHER  LICENSES  195 

33.  2240  Ibs.  of  chalk  occupy  15.5  cubic  ft.     What  is  the 
specific  gravity  of  the  chalk  ? 

NOTE.  The  specific  gravity  of  a  substance  is  its  weight  divided  by 
the  weight  of  an  equal  volume  of  water. 

34.  Find  the  value  of  the  circulating  decimal  .426  and  ex- 
plain the  solution. 

NOTE.  A  circulating  (moving  in  a  circle)  decimal  is  a  decimal 
whose  last  figures  repeat  without  end.  Dots  are  placed  over  the  first 
and  last  figures  of  the  part  which  repeats  (the  repetend).  It  is  equal 
to  a  decimal  ending  in  a  common  fraction  whose  numerator  is  the 
repetend  and  whose  denominator  is  as  many  9's  as  there  are  figures 
in  the  repetend.  Thus,  .426  =  .4||  =  f^. 


35.   Simplify:    (a)     I"*""*  °f  * 


NOTE.     See  §  122.     -s-  2^  -f-  6  is  ambiguous.    Use  the  signs  as  they 
occur. 


INDEX 


Abstract  nos.,  110. 

Addition  —  develop.,  64 ;  checks,  68 ; 
fractions,  98 ;  decimals,  105  ;  prob- 
lems, 23,  34;  signs,  132. 

Algebra  —  develop.,  131;  problems, 
38  ;  percentage,  141 ;  interest,  164  ; 
tests,  181. 

Aliquot  parts  —  multiplication,  81 ; 
div.,  88;  interest,  156;  tests,  187. 

Altitudes,  119. 

Analysis  —  develop.,  26 ;  percentage, 
142;  interest,  163;  tests,  187. 

Analytic  aid  —  develop.,  30;  int., 
163. 

Angles,  117. 

Annual  interest,  165. 

Apperception,  181. 

Arabic  notation,  55. 

Arrangement  —  prob.,  34;  para- 
graphs, 53. 

Associative  law,  191. 

Austrian  method  —  sub.,  72,  75 ; 
division,  87. 

Bank       discount  —  develop.,       158; 

tests,  192. 
Bonds  —  develop.,  160 ;  tests,  188. 

Cancellation  —  expl.,  100 ;  tests,  181. 

Canon  of  agreement,  14. 

Cases  —  develop.,  6 ;    percentage,  7  ; 

int.,  164. 
Checks,  162. 
Circle  —  cir.,    12,    121 ;     area,    121 ; 

tests,  192. 
Classifications,  2. 
Combinations  —  add.,  65;    sub.,  72; 

mult.,  77;   div.,  83. 
Commercial  discount,  147. 
Commission,  146. 
Commutative  law,  81,  100,  191. 
Complete  method,  20. 
Complex  fractions,  102,  195. 
Complex  problems,  30. 
Composite  nos.,  90. 
Compound  fractions,  102. 
Compound  interest,  165. 
Compound  numbers,  110. 


Concrete  nos.,  110. 

Cones,  119. 

Constructions,  120. 

Convex  surface,  122.  .     , 

Crutches,  53. 

Cube  root,  127,  128. 

Cylinders,  119. 

Decimals  —  develop.,    103;    history, 

109 ;  roots,  129 ;  tests,  182. 
Deduction,  16. 
Definitions,  3. 

Denominate  nos.,  110,  114,  182. 
Development  exercises,  47,  182,  188. 
Diagrams,  5,  182. 
Differences  or  differentia,  3. 
Difficult  problems,  145,  146,  194. 
Dimensions,  118. 
Discount  —  bank,  158 ;    commercial, 

147. 

Distributive  law,  191. 
Divisibility,  15,  92,  193. 
Division  —  develop.,    82;    fractions, 

99;      dec.,     105;      Austrian,     87; 

remainders,    101 ;    by  factors,  87, 

101,  108 ;   signs,  133. 
Drafts,  161,  188. 
Drill  exercises,  47,  183. 

Elementary   schools  —  scope  of   the 

grades,  169-179. 
English  notation,  62. 
Equations,  134,  192. 
Eratosthenes,  90. 
Errors,  52,  53,  183. 
Euclidean  method,  95,  193. 
Evolution,  125. 
Exact  interest,  158,  159,  178. 
Examinations  —  el.      schools,      169 ; 

teachers,  181-195. 
Exercises,  169-195. 
Explanations,  29. 
Expression,  51. 

Factors  —  develop.,  90-95 ;    division 

by,  87,  108,  101 ;  roots,  127. 
Formal  steps  of  Herbartians,  181. 
Formula,  39,  128,  141,  164. 


196 


INDEX 


197 


Fractions  —  develop.,  96 ;  lowest 
terms,  95,  98 ;  remainders,  101 ; 
roots  of,  129 ;  clearing  of,  135 ; 
g.  c.  d.  and  1.  c.  m.,  193. 

French  notation,  57. 

Fundamentals,  190. 

Games,  184. 

Geometry,  184.    See  mensuration. 

Graphic  aids,  32,  183. 

Greatest  common  divisor  —  by  fac- 
toring, 94 ;  Euclidean,  95,  193  ;  of 
fractions,  193. 

Grube  method,  189. 

Helps,  184. 
Herbartians,  181. 
Hindu  notation,  57,  62. 
Hypotenuse,  124. 

Induction,  14,  184. 

Interest  —  develop.,  152;  exact,  158, 

159,  178 ;   tests,  185,  189. 
Involution,  125. 
Italian  subtraction,  73,  75. 

Least  common  multiple  —  by  in- 
spection, 93  ;  by  factoring,  94 ; 
of  fractions,  193. 

Ledger  balances,  74. 

Lesson  plans,  46. 

Licenses  —  primary,  181 ;  higher,  190. 

Lines,  117. 

Logarithms,  129. 

Logarition,  125. 

Logical  definitions,  3,  185. 

Logical  division,  3,  185. 

Logical  steps,  9. 

Longitude  and  time,  185. 

McLellan  and  Dewey  method,  191. 

Major  analysis,  27. 

Measurements,  8. 

Mechanical  aids,  11. 

Mensuration,  117,  184. 

Mental  and  written,  50. 

Metric  system,  114. 

Minor  analysis,  27. 

Mixed  nos.,  100. 

Model  analysis,  27,  33,  187. 

Mortgages,  161. 

Multiplication  —  develop.,  76;  frac- 
tions, 16,  99;  dec.,  17,  106; 
proofs,  80 ;  signs,  132. 

Names,  3. 
Needs,  2. 


Nines,  proofs  by  —  add.,  69;   mult., 

80 ;   div.,  86. 
Notation  and  numeration  —  Arabic, 

55;      French,    57;      English,    62; 

Hindu,  63. 
Notes,  152. 
Number,  55,  190. 

Operations  combined,  88. 

Paper  folding,  11. 

Paragraphing,  53. 

Parallelograms,  5,  12,  119,  121. 

Parentheses,  88,  134. 

Partition,  22,  76,  82. 

Percentage  —  develop.,  137;  cases, 
7;  tests,  186. 

Polygons,  118. 

Postal  savings  system,  159. 

Prime  nos.,  90. 

Prisms,  119. 

Problems  —  class,  22 ;  by  experi- 
ment, 22;  analysis,  26,  30; 
algebra,  38 ;  formula,  39 ;  rule, 
40 ;  proportion,  41 ;  variation,  44 ; 
arrangement,  34 ;  proofs,  36 ; 
graphic  aids,  32 ;  explanations,  29  ; 
exercises,  169-195. 

Profit  and  loss,  144. 

Proofs  —  problems,  36:  add.,  68; 
sub.,  73;  mult.,  80;  div.,  86; 
tests,  186,  188. 

Proportion,  41,  186. 

Pyramids,  119. 

Quadrilaterals,  5,  121. 
Quotition,  22,  76,  82. 

Ratio,  43,  186,  190,  191. 
Records,  54. 
Rectangles,  5,  110,  121. 
Regular  polygons,  118. 
Remainders,  101. 
Rhomboid  and  rhombus,  5. 
Right-angle  triangle,  118,  124 
Roman  notation,  60. 
Rules,  40,  141,  164. 

Savings  banks,  167. 
Sequence  of  signs,  88,  195. 
Short  processes,  192. 
Sieve  of  Eratosthenes,  90. 
Signs,  88,  132,  133. 
Similarity,  45,  124. 
Simple  problems,  22,  26,  34. 
Solids,  117. 


198 


INDEX 


Species,  2. 

Specific  gravity,  194. 
Speer  method,  190. 
Sphere,  13,  119. 
Spiral  method,  189. 
Square  root,  127. 
Stocks,  148. 

Subtraction  —  develop.,  64,  71 ;  frac- 
tions, 98;   dec.,  105;  signs,  133. 
Surface,  117. 
Syllogism,  193. 


Tetrahedron,  119. 
Textbook,  186. 
Theory,  189. 
Transposing,  134. 
Trapezium,  5. 
Trapezoid,  5,  119,  121. 
Triangles,  4,  11,  12,  118,  124. 

Unit  of  measure,  186. 
Unitary  analysis,  191. 

Variation,  44. 


CONSULTATION  BY  CORRESPONDENCE 

For  Superintendents,  Principals,  Teachers 
and  Others 

By  Middlesex  A.  Bailey,  Author  of  "  Teaching  Arithmetic " 


ORDINARY  CONSULTATION,  $1.00 

Special  Consultation  Proportional  to  the  Amount  of  Labor 
PAYMENT  IN  ADVANCE 

THE  establishment  of  a  bureau  of  consultation  for  the 
teaching  of  arithmetic  is  something  of  an  innovation. 
In  other  lines,  laymen  consult  physicians,  and  physicians  con- 
sult specialists;  contractors  consult  engineers,  and  engineers 
consult  specialists ;  and  so  on. 

Occasions  arise  when  a  superintendent  wishes  to  consult  in 
regard  to  a  course  of  study ;  a  principal,  in  regard  to  grade 
work ;  a  teacher,,  in  regard  to  class  work ;  or  a  prospective 
teacher,  in  regard  to  preparation. 

It  is  with  diffidence  that  the  writer  offers  his  services,  be- 
cause he  realizes  his  deficiencies.  It  is  in  place,  however,  to 
state  his  preparation.  Since  graduation  from  college  in  1877, 
he  has  been  engaged  in  school  work ;  3  years  as  principal  of 
an  elementary  school  in  Winsted,  Conn. ;  5  years  as  princi- 
pal of  a  high  school  in  Keene,  N.H. ;  14  years  as  head  of  the 
department  of  mathematics  at  the  State  Normal  School  of 
Kansas,  in  Emporia ;  and  14  years  as  head  of  the  depart- 
ment of  mathematics  at  the  New  York  Training  School  for 
Teachers  in  New  York  City.  During  these  36  years  he  has 
made  the  subject  of  teaching  arithmetic  his  major  study  and 
has  written  a  series  of  arithmetics  published  by  the  American 
Book  Company. 

MIDDLESEX    A.    BAILEY,    YONKERS,    NEW    YORK 


INSTRUCTION  BY  CORRESPONDENCE 

For  Teachers  and  Prospective  Teachers  of 
Arithmetic 

By  Middlesex  A.  Bailey,  Author  of  " Teaching  Arithmetic" 


THIRTY  LESSONS,  $15.00 

Book,  Paper,  Envelopes,  Postage  both  ways,  Included 
PAYMENT  IN  ADVANCE 

A  COURSE  of  training  in  methods  of  teaching  arithmetic 
is  offered  to  those  who  are  preparing  to  obtain  a  primary 
license. 

Any  person  seventeen  years  of  age  who  has  mastered  the 
requirements  in  mathematics  of  an  ejght-year  course  in  the 
elementary  schools  is  eligible  to  take  this  work.  A  thoroughly 
satisfactory  preparation  is  impossible  without  some  knowl- 
edge of  algebra  and  geometry.  If  the  candidate  has  not  taken 
these  subjects,  he  should  arrange  to  study  them  by  correspond- 
ence or  otherwise  at  the  earliest  opportunity.  He  is  advised 
also  to  purchase  Jevon  &  Hill's  Logic  and  to  read  it  in  con- 
nection with  the  study  of  methods.  It  should  be  borne  in 
mind  that  no  course  by  correspondence  can  equal  in  value 
attendance  at  a  normal  or  training  school. 

A  second  course  is  offered  to  teachers  who  are  preparing  for 
a  higher  license  and  who  prefer  work  by  correspondence  to 
attendance  in  summer  schools  or  afternoon  classes. 

The  candidate  is  requested  to  state  in  advance  his  age,  the 
schools  from  which  he  has  graduated,  the  schools  which  he 
has  attended,  the  purpose  for  which  he  wishes  to  take  the 
work  and  the  course  desired.  The  candidate  for  a  higher 
license  is  requested  also  to  state  his  teaching  experience. 

MIDDLESEX   A.    BAILEY,   YONKERS,    NEW    YORK 


UNIVERSITY  OF  CALIFORNIA  AT  LOS  ANGELES 

THE  UNIVERSITY  LIBRARY 
This  book  is  DUE  on  the  last  date  stamped  below 


WOV  9 


3J8 


19*1 


3 


NOV      5  1946 


APR  1  4  1947 


nc^ 

teC  13  1950< 
FEB-2  1952 

Form  L-9-15m-3,'34 


8 


,  JAN 


19513 


If  AY  a  6  1951 

<    l»  1  1  1954 
"MAY  101956 


APR  2  2  195S 


DEC  9 

REC'D  COL 


UNIVERSITY  of  CALIFORNIA 


JUBRABT 


